The post TRAVEL TO HARIDWAR RISHIKESH NEELKANTH appeared first on Learn Online.

]]>People want to travel everywhere. Before travelling arises many questions about how to go, stay and important question about budget. Most of families belongs to middle class so they want to cover more area in less budget. So keep in mind all these things I am going to share my experiences travel to HARIDWAR RISHIKESH NEELKANTH and Important places to watch for viewers.

We are two families plan for tour to haridwar rishikesh and neekanth. My colleague was very experienced for this tour so its very easy for me. I was very free to fear about where to go and how to go , budget etc.

We plan for three days Friday, Saturday and Sunday. So there are many trains go to HARIDWAR. We decided go to early in the morning around 2.00 A.M o’clock early in the morning on Friday. We reached HARIDWAR same day on Friday around 7 o’clock A.M. So we have enough time to go anywhere. We take auto from Railway station to Bhimgoda.

Because our stay was near to Bhimgoda street in known Ashram. Bhimgoda is good to stay due to very close to har ki pauri. Approximately 10 mint walking distance from here for Har ki pauri Ganga. After taking rest and refresh we ready to go. First we decided to go for Mansa Devi by Timber Train. But there was very rush. So we go beck from here. After some walking we hire auto for Chandi Devi at good price. From here we take package to up down by Timber Train for Chandi Devi darshan. It was amazing to travel by Timber Train. There was heavy rush and long waiting time to travel by timber train so we spent almost whole day there. In the evening go back to Har ki pauri for Ganga Aarti. It was amazing to see Ganga aarti at har ki pauri. After aarti we were very tired and go back to ashram.

Next day Morning we plan to go Rishikesh. We hire common auto for Rishikesh from Kharkhri chownk near bhimgoda. First we go to Lakshman Jhoola. We take breakfast here and watch Lakshman Jhoola. From here jeeps are going to Neelkanth (Jeeps for Neelkanth are going from both Ram Jhoola and Lakshman Jhoola) . We take 4 tickets for neelkanth. We travel for neelkanth with other travelers. After Neelkanth darshan we go back to Lakshman Jhoola and get another jeep for Ram jhoola. Here many temples to see in Rishikesh in which are like: Parmath ashram, Shanti Kunj etc. Also enjoy in boat from one side to other side. You can also do Rafting in Rishikesh. So in the evening we come back to Haridwar in Ashram.

Next morning around 4 o’clock first we ready to go Har ki Pauri Ganga. We take sanaan( Bath) in Ganga which is owsome and very relief to body. After come back to Ashram take breafast with Kachori which is very tasty. And now ready to go for Mansa devi again in the early morning. Now this time was not rush due to working day. We easily go by timber train to Maa Mansa Devi. A total package of Maa Mansa Devi and Chandi devi collectively is expensive than separate package for both differently. After this we go for others famous temples of Haridwar: Maa Vaishno Devi Temple, Pawan Dham or Sheesh Mahal, Dudharu Temple etc. Many temple are visit to here in Haridwar. So in the evening we come back to Ashram and ready to go for station because now the time come when we are ready to coming back home after wonderful journey. Out train is already booked in the night. So in the morning around 3am we reach at our home. We all families were very happy after this wonderful journey.

Brief Discussion: Haridwar – Har Ki Pauri, Maa Mansa Devi, Maa Chandi Devi by Timber train must go. Other temples like Maa Vaishno Devi, pwan Dham etc.

Rishikesh: Go to Neelkanth from Lakshman Jhoola, Ram Jhoola, Parmarth Ashram, shanti kunj etc

I hope my this blog will help you alot. Please comment if you want to know anything.

Thanks

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]]>The post Tour To Agra-Mathura-Varindavan appeared first on Learn Online.

]]>Hello friends today I am going to share my experience for tour to Agra-Mathura-Vaindavan. First of all I will suggest you the best time to travel for this. You can travel here from mid october to March. I travel starting of the month October but there is little bit hot. You can plan at these places for four to five days. In these days you can easily watch all famous places of these area or complete your 84 koss yatra.

We most probably travel into night and in the morning reach at our destination. So decided to travel at friday night(29-sep-17). Firstly we decided to go Agra. So I booked railway tickets two months and book room in hotel 40 days ago in Agra. We are two families to travel. You can enjoy more if you go together and less costly. Agra is approximately 500 KM from out home town.

The day come when we are ready to go friday night. But unfortunately in evening we check that the train has been cancelled for some reasons. So we both families decided to go by bus due to no train come. We reach at bus stoppage and take bus for New Delhi. In this time we check that train is coming but 4-5 hours late and we already file online TDR at irctc website for “Train is late more than 3 hours and Passengers not travelling” after waiting 4 hours.

In the morning 3 am we reach at new delhi. The buses for Mathura or Agra from Sarai Kale Khan. So we decided to rest at new delhi bus stand due to very tired and our hotel check in was 12 PM. After take rest and fresh at Delhi bus stand we decided to go by Metro Train upto Old Faridabad which is good if you want travel by metro otherwise I suggest go directly by train or bus from Sarai Kale Khan. We take another bus from faridabad to mathura and mathura to Agra. So we reach Agra next day (30-sep-17) near about 12:15 o’clock. From here we take auto to our booked 3 star hotel which is deny to free room due to heavy rush or shift somewhere else we discuss it later, so in the end we stay the same online booked hotel on some conditions which is comfortable for us.

After taking rest for two ours 1 to 3 o’clock we ready to go Taj Mahal. Firstly you have to buy entry ticket at counter inside taj mahal and then go in the line to enter in the taj mahal. After see Taj Mahal we come back to hotel due to tired. You can also go to Red Fort, Jama Masjid,Mahtab Bagh etc in agra or Go to Fatehpur sikri on next day.

Next day(1-OCT-17) in the morning around 7 o’clock we leave the hotel and go for Mathura. We reach at agra Railway station and take tickets for Mathura. During this time breakfast at railway station. After reach at 10:30 clock Mathura we decided to go first at Shri Krishna Janam Bhumi. We take e-rickshaw in 80rs for Shri Krishna Janam Bhumi. Here Mobile phone, luggage is not allowed to enter in the temple. So you have to submit there. We enter by second gate into the temple and one of us stay outside to keep luggage and mobile phones. After see temple we ready to go Varindavan. Lassi of curd is very tasty in Mathura. Must drink here. Distance from Mathura to Varindavan is 12 km where we want to go near Bankey Bihari Mandir. I suggest to take room near bankey bihari mandir.

Now we reach at Varindavan. There was heavy rush in the Varindavan due to Krishna ikadshi, weekend and holidays. We surprised after looking heave rush. Because we decided to take any guest house at the sport. There are many ashram and guest house in Varindavan but we afraid after looking so much crowed. It was our first time visit in the Varindavan. We dont understand where to go. No one ashram guru known by us. So we take relaxed somewhere else and kept our luggage. So search some rooms here. Due to heave rush and new to here people demanding so much much money for simple and unorganised rooms in their houses. So after search at Bankey bihar road we found well furnished and good AC room. Normal room upto 600rs and AC rooms are upto 800. But in heavey rush AC room upto 1000rs single room. Two family can easily stay in single room with extra Sleeping mattress.

So in the evening first we go “Prem Mandir” near about 4 o’clock. Prem mandir is 2 km from bankey bihari mandir. You can go by rickshaw in 40rs(4 Members). Prem Mandir is very attractive temple. So many lightening in temple feel different. After visit Prem mandir we go Iskon mandir by walking which is around 600 meter from Prem Mandir. You can shop here for Kangan at good price. We go to in the room and after refresh we go for dinner. You can take dinner as Thali system.

Next Morning(2-Oct-17) I wake up early in the morning 4 o’clock and ready upto 5 oclock. And wake up others until they are going to ready I go to “Radha Vallabh” mandir to attend Krishna Aarti. Bankey Bihari Aarti start 7:30 am around, First Aarti Radha Vallabh at 5 o’clock. After aarti mandir closed upto 8:30 for Shingar. And reopen after shingar of Krishna. So after taking tea all are ready to go near about 6:40. Fisrt we again go to Radha Vallabh which is closed. Some guide come and offer to saw Nidhi van, old mandir on just rs 50. So we ready to go with him by walking behind him. He was very good for us because we saw all near mandir and Nidhi Van. He also told us story of all these historical places. We reach at Jamuna ji near Kausi ghaat. We hire a private vote at 200rs for a full circle around upto Kausi Ghaat. Now we go back to Bankey bihari mandir at erikshaw on just rs 10 for each. There was very crowed.We go inside temple and worship Bankey Bihari. After Bankey bihari we come back to guest house. Now the time has been 10 o’clock around. We decided to go in hurry for Govind Dev Temple and Ragnath Temple. We hire again erikshaw at 80 rs for Govind Dev temple. It was very old and nice temple. By walikng from here we go to raagnath temple just 200mtrs away from it. It was very beautiful temple. We worship in both temples and closed at 12 oclock. Its remember that all temples in Mathura anc Varindavan are closex from 12 pm to 4 pm, so try to go early in thr morning or late evening. So after taking lunch we go back to guest house and ready to go Mathura railway station. So It was my all experience to share with all devotees and traveller. Hope it will helpfull to you. Thnkas for reading my blog. Raaadhey Raaadhey

We most probably travel into night and in the morning reach at our destination. So decided to travel at friday night(29-sep-17). Firstly we decided to go Agra. So I booked railway tickets two months and book room in hotel 40 days ago in Agra. We are two families to travel. You can enjoy more if you go together and less costly. Agra is approximately 500 KM from out home town.

The day come when we are ready to go friday night. But unfortunately in evening we check that the train has been cancelled for some reasons. So we both families decided to go by bus due to no train come. We reach at bus stoppage and take bus for New Delhi. In this time we check that train is coming but 4-5 hours late and we already file online TDR at irctc website for “Train is late more than 3 hours and Passengers not travelling” after waiting 4 hours.

In the morning 3 am we reach at new delhi. The buses for Mathura or Agra from Sarai Kale Khan. So we decided to rest at new delhi bus stand due to very tired and our hotel check in was 12 PM. After take rest and fresh at Delhi bus stand we decided to go by Metro Train upto Old Faridabad which is good if you want travel by metro otherwise I suggest go directly by train or bus from Sarai Kale Khan. We take another bus from faridabad to mathura and mathura to Agra. So we reach Agra next day (30-sep-17) near about 12:15 o’clock. From here we take auto to our booked 3 star hotel which is deny to free room due to heavy rush or shift somewhere else we discuss it later, so in the end we stay the same online booked hotel on some conditions which is comfortable for us.

After taking rest for two ours 1 to 3 o’clock we ready to go Taj Mahal. Firstly you have to buy entry ticket at counter inside taj mahal and then go in the line to enter in the taj mahal. After see Taj Mahal we come back to hotel due to tired. You can also go to Red Fort, Jama Masjid,Mahtab Bagh etc in agra or Go to Fatehpur sikri on next day.

Next day(1-OCT-17) in the morning around 7 o’clock we leave the hotel and go for Mathura. We reach at agra Railway station and take tickets for Mathura. During this time breakfast at railway station. After reach at 10:30 clock Mathura we decided to go first at Shri Krishna Janam Bhumi. We take e-rickshaw in 80rs for Shri Krishna Janam Bhumi. Here Mobile phone, luggage is not allowed to enter in the temple. So you have to submit there. We enter by second gate into the temple and one of us stay outside to keep luggage and mobile phones. After see temple we ready to go Varindavan. Lassi of curd is very tasty in Mathura. Must drink here. Distance from Mathura to Varindavan is 12 km where we want to go near Bankey Bihari Mandir. I suggest to take room near bankey bihari mandir.

Now we reach at Varindavan. There was heavy rush in the Varindavan due to Krishna ikadshi, weekend and holidays. We surprised after looking heave rush. Because we decided to take any guest house at the sport. There are many ashram and guest house in Varindavan but we afraid after looking so much crowed. It was our first time visit in the Varindavan. We dont understand where to go. No one ashram guru known by us. So we take relaxed somewhere else and kept our luggage. So search some rooms here. Due to heave rush and new to here people demanding so much much money for simple and unorganised rooms in their houses. So after search at Bankey bihar road we found well furnished and good AC room. Normal room upto 600rs and AC rooms are upto 800. But in heavey rush AC room upto 1000rs single room. Two family can easily stay in single room with extra Sleeping mattress.

So in the evening first we go “Prem Mandir” near about 4 o’clock. Prem mandir is 2 km from bankey bihari mandir. You can go by rickshaw in 40rs(4 Members). Prem Mandir is very attractive temple. So many lightening in temple feel different. After visit Prem mandir we go Iskon mandir by walking which is around 600 meter from Prem Mandir. You can shop here for Kangan at good price. We go to in the room and after refresh we go for dinner. You can take dinner as Thali system.

Next Morning(2-Oct-17) I wake up early in the morning 4 o’clock and ready upto 5 oclock. And wake up others until they are going to ready I go to “Radha Vallabh” mandir to attend Krishna Aarti. Bankey Bihari Aarti start 7:30 am around, First Aarti Radha Vallabh at 5 o’clock. After aarti mandir closed upto 8:30 for Shingar. And reopen after shingar of Krishna. So after taking tea all are ready to go near about 6:40. Fisrt we again go to Radha Vallabh which is closed. Some guide come and offer to saw Nidhi van, old mandir on just rs 50. So we ready to go with him by walking behind him. He was very good for us because we saw all near mandir and Nidhi Van. He also told us story of all these historical places. We reach at Jamuna ji near Kausi ghaat. We hire a private vote at 200rs for a full circle around upto Kausi Ghaat. Now we go back to Bankey bihari mandir at erikshaw on just rs 10 for each. There was very crowed.We go inside temple and worship Bankey Bihari. After Bankey bihari we come back to guest house. Now the time has been 10 o’clock around. We decided to go in hurry for Govind Dev Temple and Ragnath Temple. We hire again erikshaw at 80 rs for Govind Dev temple. It was very old and nice temple. By walikng from here we go to raagnath temple just 200mtrs away from it. It was very beautiful temple. We worship in both temples and closed at 12 oclock. Its remember that all temples in Mathura anc Varindavan are closex from 12 pm to 4 pm, so try to go early in thr morning or late evening. So after taking lunch we go back to guest house and ready to go Mathura railway station. So It was my all experience to share with all devotees and traveller. Hope it will helpfull to you. Thnkas for reading my blog. Raaadhey Raaadhey

In Brief:

Mathura: Shri Krishna Janam Bhumi, Dwarkadheesh, Radha Kund, Gukul, Barsana, Nand Gaon

Varindavan: Shri Bankey Bihrai Temple, Radha Vallabh, Prem Mandir, Iskon Mandir, Govind Dev Temple, Raagnath Temple

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]]>The post Add Table into WordPress and Justify Text appeared first on Learn Online.

]]>In this post I will explain how to add Table and how to remove border of table into wordpress blog. The most important is how to justify text content of table.

Let first discuss how to add Table into WordPress

Simple in Text tab type <Table> and close table </table>.

<table> /** Table starting symbol **/

<tbody> /** not compulsary –Table body symbol **/

<tr> /* creates rows with in table **/

<td> /** creates cell within a row **/

</td>

</tr>

</tbody>

</table>

**For Example: Divide table row into fixed different column size into WordPress HTML**

<table style=”height: 154px;” border=”none” width=”765″>

<tbody>

<tr>

<td width=”16%”>

<div style=”text-align: left;”>

<span style=”font-size: 14px;”>SUBJECT</a></span></div></td>

<td width=”60%”>

<div style=”text-align: left;”>

<span style=”font-size: 14px; text-align: justify; float: left;”> This is about how to align text content into proper end </span></div></td>

<td width=”15%”> text

</td>

</tr>

</tbody>

**How To** **Justify Text content at proper end**

In above example shows that how to align properly. When we type text at the end of the line word not properly comes as above line. So use below code to justify the line end as proper in above line.

**text-align: justify; float: left;**

<span style=”font-size: 14px; text-align: justify; float: left;”>

**How To remove Border Lines of Table**

To remove border line you have to update style.css file.

Move cursor on **Appearance **and select **Editor**

Find by “Ctrl+ F” table and page down you will find caption and update below

td { border-width: 0 0px 1px 0;} “Update Pixel as width of table border you want”

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]]>The post North India appeared first on Learn Online.

]]>The post Xpert Solution For class 10 chapter 3: Pair of Linear Equations in Two Variables appeared first on Learn Online.

]]>- Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically.

**Solution:**

Let present age of Aftab be *x*

And, present age of daughter is represented by *y*

Then Seven years ago,

Age of Aftab = *x *-7

Age of daughter = *y*-7

According to the question,

(*x *– 7) = 7 (*y *– 7 )

*x *– 7 = 7 *y *– 49

*x*– 7*y *= – 49 + 7

*x *– 7*y* = – 42 …**(i)**

*x* = 7*y* – 42

Putting *y* = 5, 6 and 7, we get

*x* = 7 × 5 – 42 = 35 – 42 = – 7

*x* = 7 × 6 – 42 = 42 – 42 = 0

*x* = 7 × 7 – 42 = 49 – 42 = 7

x |
-7 | 0 | 7 |

y |
5 | 6 | 7 |

Three years from now ,

Age of Aftab = *x *+3

Age of daughter = *y *+3

According to the question,

(*x *+ 3) = 3 (*y *+ 3)

*x *+ 3 = 3*y *+ 9

*x *-3*y *= 9-3

*x *-3*y *= 6 …**(ii)**

*x *= 3*y *+ 6

Putting, *y *= -2,-1 and 0, we get

*x *= 3 × – 2 + 6 = -6 + 6 =0

*x *= 3 × – 1 + 6 = -3 + 6 = 3

*x *= 3 × 0 + 6 = 0 + 6 = 6

x |
0 | 3 | 6 |

y |
-2 | -1 | 0 |

Algebraic representation

From equation **(i)** and **(ii)**

*x *– 7*y *= – 42 …**(i)**

*x *– 3*y *= 6 …**(ii)**

Graphical representation

- The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys another bat and 3 more balls of the same kind for Rs 1300. Represent this situation algebraically and geometrically.

**Solution:**

Let cost of one bat = Rs *x*

Cost of one ball = Rs *y*

3 bats and 6 balls for Rs 3900 So that

3*x *+ 6*y* = 3900 … **(i)**

Dividing equation by 3, we get

*x *+ 2*y* = 1300

Subtracting 2*y* both side we get

*x* = 1300 – 2*y *

Putting *y* = -1300, 0 and 1300 we get

*x* = 1300 – 2 (-1300) = 1300 + 2600 = 3900

*x *= 1300 -2(0) = 1300 – 0 = 1300

*x* = 1300 – 2(1300) = 1300 – 2600 = – 1300

x |
3900 | 1300 | -1300 |

y |
-1300 | 0 | 1300 |

Given that she buys another bat and 2 more balls of the same kind for Rs 1300

So, we get

*x* + 2*y *= 1300 … **(ii)**

Subtracting 2y both side we get

*x *= 1300 – 2*y*

Putting *y* = – 1300, 0 and 1300 we get

*x* = 1300 – 2 (-1300) = 1300 + 2600 = 3900

*x *= 1300 – 2 (0) = 1300 – 0 = 1300

*x *= 1300 – 2(1300) = 1300 – 2600 = -1300

x |
3900 | 1300 | -1300 |

y |
-1300 | 0 | 1300 |

Algebraic representation

3*x *+ 6*y* = 3900 … **(i)**

*x *+ 2*y *= 1300 … **(ii)**

Graphical representation,

- The cost of 2 kg of apples and 1kg of grapes on a day was found to be Rs 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs 300. Represent the situation algebraically and geometrically.

**Solution:**

Let cost each kg of apples = Rs *x*

Cost of each kg of grapes = Rs *y*

Given that the cost of 2 kg of apples and 1kg of grapes on a day was found to be Rs 160

So that

2 *x *+ *y *= 160 … **(i)**

2*x* = 160 – *y*

*x* = (160 – *y*)/2

Let *y* = 0 , 80 and 160, we get

*x* = (160 – ( 0 )/2 = 80

*x* = (160- 80 )/2 = 40

*x* = (160 – 2 × 80)/2 = 0

x |
80 | 40 | 0 |

y |
0 | 80 | 160 |

Given that the cost of 4 kg of apples and 2 kg of grapes is Rs 300

So we get

4*x* + 2*y *= 300 … **(ii)**

Dividing by 2 we get

2*x* + *y* = 150

Subtracting 2*x* both side, we get

*y *= 150 – 2*x*

Putting *x* = 0 , 50 , 100 we get

*y *= 150 – 2 × 0 = 150

*y *= 150 – 2 × 50 = 50

*y *= 150 – 2 × (100) = -50

x |
0 | 50 | 100 |

y |
150 | 50 | -50 |

Algebraic representation,

2*x *+ *y* = 160 … **(i)**

4*x* + 2*y* = 300 … **(ii)**

Graphical representation,

** ****Exercise 3.2**

**Form the pair of linear equations in the following problems, and find their solutions graphically.**

(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.

**Solution:**

Let number of boys = *x*

Number of girls = *y*

Given that total number of student is 10 so that

*x *+ *y *= 10

Subtract y both side we get

*x *= 10 – *y*

Putting *y *= 0 , 5, 10 we get

*x *= 10 – 0 = 10

*x *= 10 – 5 = 5

*x *= 10 – 10 = 0

x |
10 | 5 |

y |
0 | 5 |

Given that If the number of girls is 4 more than the number of boys

So that

*y *= *x *+ 4

Putting *x* = -4, 0, 4, and we get

*y *= – 4 + 4 = 0

*y *= 0 + 4 = 4

*y *= 4 + 4 = 8

x |
-4 | 0 | 4 |

y |
0 | 4 | 8 |

Graphical representation

Therefore, number of boys = 3 and number of girls = 7.

**(ii)** 5 pencils and 7 pens together cost Rs 50, whereas 7 pencils and 5 pens together cost Rs 46. Find the cost of one pencil and that of one pen.

**Solution:**

Let cost of pencil = Rs *x*

Cost of pens = Rs *y*

5 pencils and 7 pens together cost Rs 50,

So we get

5*x* + 7*y* = 50

Subtracting 7*y* both sides we get

5*x* = 50 – 7*y*

Dividing by 5 we get

*x* = 10 – 7 y /5

Putting value of *y* = 5 , 10 and 15 we get

*x* = 10 – 7 × 5/5 = 10 – 7 = 3

*x* = 10 – 7 × 10/5 = 10 – 14 = – 4

*x* = 10 – 7 × 15/5 = 10 – 21 = – 11

x |
3 | -4 | -11 |

y |
5 | 10 | 15 |

Given that 7 pencils and 5 pens together cost Rs 46

7*x* + 5*y* = 46

Subtracting 7*x* both side we get

5*y* = 46 – 7*x*

Dividing by 5 we get

*y* = 46/5 – 7*x*/5

y = 9.2 – 1.4*x*

Putting *x* = 0 , 2 and 4 we get

*y* = 9.2 – 1.4 × 0 = 9.2 – 0 = 9.2

*y* = 9.2 – 1.4 (2) = 9.2 – 2.8 = 6.4

*y* = 9.2 – 1.4 (4) = 9.2 – 5.6 = 3.6

x |
0 | 2 | 4 |

y |
9.2 | 6.4 | 3.6 |

Graphical representation

Therefore, cost of one pencil = Rs 3 and cost of one pen = Rs 5.

- On comparing the ratios
*a**₁*/*a**₂*,*b**₁*/*b₂*and*C**₁*/*c₂*, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident.

**Solution:**

(i) 5*x* – 4*y* + 8 = 0

7*x* + 6*y* – 9 = 0

Comparing these equation with

*a**₁**x* + *b**₁**y* + *C**₁* = 0

*a**₂**x* + *b**₂**y* + *C**₂*= 0

We get

*a**₁* = 5, *b**₁*= -4, and *C**₁* = 8

*a**₂*=7, *b₂* = 6 and C₂= -9

*a₁/a₂ = 5/7,
b₁/ b₂ = -4/6 and
C₁/C₂= 8/-9
Hence, a₁/a₂ ≠ b₁/b₂*

Therefore, both are intersecting lines at one point.

(ii) 9*x* + 3*y* + 12 = 0

18*x* + 6*y* + 24 = 0

Comparing these equations with

*a**₁**x* + *b**₁**y* + *C**₁* = 0

*a**₂**x* + *b**₂**y* + *C**₂*= 0

We get

*a**₁* = 9, *b**₁*= 3, and *C**₁* = 12

*a**₂* = 18, *b₂* = 6 and *C**₂*= 24

*a**₁*/*a**₂* = 9/18 = 1/2

*b**₁*/*b₂* = 3/6 = 1/2 and

*C**₁*/*C**₂*= 12/24 = 1/2

Hence, *a**₁*/*a**₂* = *b**₁*/*b₂*_{ }=_{ }*C**₁*/C*₂*

Therefore, both lines are coincident

(iii) 6*x* – 3*y* + 10 = 0

2*x* – *y* + 9 = 0

Comparing these equations with

*a**₁**x* + *b**₁**y* + *c**₁* = 0

*a**₂**x* + *b**₂**y* + *c**₂*= 0

We get

*a**₁* = 6, *b**₁*= -3, and *C**₁* = 10

*a**₂* = 2, *b₂* = -1 and C₂= 9

*a**₁*/*a**₂* = 6/2 = 3/1

*b**₁*/*b₂* = -3/-1 = 3/1 and

*C**₁*/*C**₂*= 12/24 = 1/2

Hence, *a**₁*/a₂ = *b**₁*/*b₂*_{ }≠_{ }*C**₁*/*c₂*

Therefore, both lines are parallel

**On comparing the ratios***a**₁***/***a**₂***,***b**₁***/***b₂***and***C**₁***/***C**₂***find out whether the following pair of linear equations are consistent, or inconsistent.**

(i) 3*x*+ 2*y*= 5 ; 2*x*– 3*y*= 7

(ii) 2*x*– 3*y*= 8 ; 4*x*– 6*y*= 9

(iii) 3/2*x*+ 5/3*y*= 7 ; 9*x*– 10*y*= 14

(iv) 5*x*– 3*y*= 11 ; – 10*x*+ 6*y*= –22

(v) 4/3*x*+ 2*y*=8 ; 2*x*+ 3*y*= 12

**Solution:**

(i) 3*x* + 2*y* = 5 ; 2*x* – 3*y* = 7

*a**₁*/*a**₂* = 3/2

*b**₁*/*b₂* = -2/3 and

*C**₁*/*C**₂*= 5/7

Hence, *a**₁*/*a**₂* ≠ b₁/*b₂*

These linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent.

(ii) 2*x* – 3*y* = 8 ; 4*x* – 6*y* = 9

*a**₁*/*a**₂* = 2/4 = 1/2

*b**₁*/*b₂* = -3/-6 = 1/2 and

*C**₁*/*C**₂*= 8/9

Hence, *a**₁*/a₂ = *b**₁*/*b*_{2 }≠_{ }*C**₁*/*c*_{2}

_{
}Therefore, these linear equations are parallel to each other and thus have no possible solution. Hence, the pair of linear equations is inconsistent.

_{
}(iii) 3/2*x* + 5/3*y* = 7 ; 9*x *– 10*y* = 14

*a**₁*/*a**₂* = 3/2/9 = 1/6

*b**₁*/*b₂* = 5/3/-10 = -1/6 and

*C**₁*/*C**₂*= 7/14 = 1/2

Hence, *a**₁*/a₂ ≠ *b**₁*/*b₂*

_{
}Therefore, these linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent.

(iv) 5*x* – 3*y* = 11 ; – 10*x* + 6*y* = –22

*a**₁*/*a**₂* = 5/-10 = -1/2

*b**₁*/*b₂ *= -3/6 = -1/2 and

*C**₁*/*C**₂*= 11/-22 = -1/2

Hence, *a**₁*/a₂ = *b**₁*/*b₂*_{ = }*C**₁*/*cb₂*

_{
}Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent.

(v) 4/3*x* + 2*y* =8 ; 2*x* + 3*y* = 12

*a**₁*/*a**₂* = 4/3/2 = 2/3

*b**₁*/*b₂ *= /3 and

*C**₁*/*C**₂*= 8/12 = 2/3

Hence, *a**₁*/a₂ = *b**₁*/*b₂*_{= }*C**₁*/*C ₂*

**Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:**

(i)*x*+*y*= 5, 2*x*+ 2*y*= 10

(ii) *x* – *y* = 8, 3*x* – 3*y* = 16

(iii) 2*x* + *y* – 6 = 0, 4*x* – 2*y* – 4 = 0

(iv) 2*x* – 2*y* – 2 = 0, 4*x* – 4*y* – 5 = 0

**Solution:**

(i)* x* + *y* = 5; 2*x* + 2*y* = 10

*a**₁*/*a**₂* = 1/2

*b**₁*/*b₂* = 1/2 and

*C**₁*/*C**₂*= 5/10 = 1/2

Hence, *a**₁*/a₂ = *b**₁*/*b₂*_{ = }*C**₁*/*c₂*

Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent.

*x* + *y* = 5

*x* = 5 – *y*

x |
4 | 3 | 2 |

y |
1 | 2 | 3 |

And, 2*x* + 2*y* = 10

*x* = 10-2y/2

x |
4 | 3 | 2 |

y |
1 | 2 | 3 |

Graphical representation

From the figure, it can be observed that these lines are overlapping each other. Therefore, infinite solutions are possible for the given pair of equations.

**(ii)** *x* – *y* = 8, 3*x* – 3*y* = 16

*a**₁*/*a**₂* = 1/3

*b**₁*/*b₂* = -1/-3 = 1/3 and

*C**₁*/*C**₂*= 8/16 = 1/2

Hence, *a**₁*/a₂ = *b**₁*/*b₂*_{ }≠_{ }*C**₁*/*C₂ _{ }*Therefore, these linear equations are parallel to each other and thus have no possible solution. Hence, the pair of linear equations is inconsistent.(iii) 2

Hence,

Therefore, these linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent.

2*x* + *y* – 6 = 0

*y* = 6 – 2*x*

x |
0 | 1 | 2 |

y |
6 | 4 | 2 |

And, 4*x* – 2*y* -4 = 0

*y* = 4*x* – 4/2

x |
1 | 2 | 3 |

y |
0 | 2 | 4 |

Graphical representation

From the figure, it can be observed that these lines are intersecting each other at the only one point i.e., (2,2) which is the solution for the given pair of equations.

(iv) 2*x* – 2*y* – 2 = 0, 4*x* – 4*y* – 5 = 0

*a**₁*/*a**₂* = 2/4 = 1/2

*b**₁*/*b₂*= -2/-4 = 1/2 and

*C**₁*/*C**₂*= 2/5

Hence, *a**₁*/a₂ = *b**₁*/*b₂*_{ }≠_{ }*C**₁*/*C ₂*

- Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

**Solution:**

Let length of rectangle = *x* m

Width of the rectangle = *y *m

According to the question,

*y* – *x* = 4 … **(i)**

*y* + *x* = 36 … **(ii)**

*y* – *x* = 4

*y* = *x* + 4

x |
8 | 12 | |

y |
4 | 12 | 16 |

*y* +* x* = 36

x |
36 | 16 | |

y |
36 | 20 |

Graphical representation

From the figure, it can be observed that these lines are intersecting each other at only point i.e., (16, 20). Therefore, the length and width of the given garden is 20 m and 16 m respectively.

- Given the linear equation 2
*x*+ 3*y*– 8 = 0, write another linear equations in two variables such that the geometrical representation of the pair so formed is:

(i) intersecting lines

(ii) parallel lines

(iii) coincident lines

**Solution:**

(i) Intersecting lines:

For this condition,

*a**₁*/*a**₂* ≠ b₁/b*₂*

The second line such that it is intersecting the given line is

2*x* + 4*y* – 6 = 0 as

*a**₁*/*a**₂* = 2/2 = 1

*b**₁*/*b₂* = 3/4 and

*a**₁*/*a**₂* ≠ b₁/b*₂*

_{
}(ii) Parallel lines

For this condition,

*a**₁*/a₂ = *b**₁*/*b₂*_{ }≠_{ }*C**₁*/*c₂*

Hence, the second line can be

4*x* + 6*y* – 8 = 0 as

*a**₁*/*a**₂* = 2/4 = 1/2

*b**₁*/*b₂* = 3/6 = 1/2 and

*C**₁*/*C**₂*= -8/-8 = 1

and *a**₁*/a₂ = *b**₁*/*b₂*_{ }≠_{ }*C**₁*/*c₂*

_{
}(iii) Coincident lines

For coincident lines,

*a**₁*/a₂ = *b**₁*/*b₂*_{= }*C**₁*/*c₂*

Hence, the second line can be

6*x* + 9*y* – 24 = 0 as

*a**₁*/*a**₂* = 2/6 = 1/3

*b**₁*/*b₂* = 3/9 = 1/3 and

*C**₁*/*C**₂*= -8/-24 = 1/3

and *a**₁*/a₂ = *b**₁*/*b₂*_{ = }*C**₁*/*c₂*

_{
}7. Draw the graphs of the equations *x* – y + 1 = 0 and 3*x* + 2*y* – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the *x*-axis, and shade the triangular region.

**Solution:**

*x* – *y* + 1 = 0

*x* = *y* – 1

x |
1 | 2 | |

y |
1 | 2 | 3 |

3*x* + 2*y* – 12 = 0

*x* = 12 – 2*y*/3

x |
4 | 2 | |

y |
3 | 6 |

Graphical representation

From the figure, it can be observed that these lines are intersecting each other at point (2, 3) and *x*-axis at ( – 1, 0) and (4, 0). Therefore, the vertices of the triangle are (2, 3), ( – 1, 0), and (4, 0).

**Exercise 3.3**

**Solve the following pair of linear equations by the substitution method.**

(i) *x *+ *y *= 14 ; *x *– *y *= 4

(ii) *s *– *t *= 3 ; *s*/3 + *t*/2 = 6

(iii) 3*x* – *y* = 3 ; 9*x* – 3*y* = 9

(iv) 0.2*x* + 0.3*y* = 1.3 ; 0.4*x* + 0.5*y* = 2.3

(v) √2*x*+ √3*y* = 0 ; √3*x* – √8*y* = 0

(vi) 3/2*x* – 5/3*y* = -2 ; *x*/3 + *y*/2 = 13/6

**Solution:**

(i) *x* + *y* = 14 … **(i)**

*x* – *y* = 4 … **(ii)**

From equation **(i)**, we get

*x* = 14 – *y* … **(iii)**

Putting this value in equation **(ii)**, we get

(14 – *y*) – y = 4

14 – 2*y* = 4

10 = 2*y*

*y *= 5 … **(iv)**

Putting this in equation **(iii)**, we get

*x *= 9

∴ *x *= 9 and *y* = 5

(ii) *s *– *t *= 3 … **(i)**

*s*/3 + *t*/2 = 6 … **(ii)**

From equation **(i)**, we get*s *= *t* + 3

Putting this value in equation **(ii)**, we get

*t*+3/3 + *t*/2 = 6

2*t* + 6 + 3t = 36

5*t* = 30

*t* = 30/5 … **(iv)**

Putting in equation **(iii)**, we obtain

*s* = 9

∴ *s* = 9, *t* = 6

(iii) 3*x* – *y *= 3 … **(i)**

9*x* – 3*y* = 9 … **(ii)**

From equation **(i)**, we get

*y* = 3*x* – 3 … **(iii)**

Putting this value in equation **(ii)**, we get

9*x* – 3(3*x* – 3) = 9

9*x* – 9*x* + 9 = 9

9 = 9

This is always true.

Hence, the given pair of equations has infinite possible solutions and the relation between these variables can be given by

*y* = 3*x* – 3

Therefore, one of its possible solutions is *x* = 1, *y* = 0.

(iv) 0.2*x* + 0.3*y* = 1.3 … **(i) **

0.4*x* + 0.5*y* = 2.3 … **(ii)**

0.2*x* + 0.3y = 1.3

Solving equation **(i)**, we get

0.2*x* = 1.3 – 0.3*y*

Dividing by 0.2, we get

*x* = 1.3/0.2 – 0.3/0.2

*x* = 6.5 – 1.5 *y* …**(iii)**

Putting the value in equation **(ii)**, we get

0.4*x* + 0.5*y* = 2.3

(6.5 – 1.5*y*) × 0.4*x* + 0.5*y* = 2.3

2.6 – 0.6*y* + 0.5*y* = 2.3

-0.1*y* = 2.3 – 2.6

*y *= -0.3/-0.1

*y* = 3

Putting this value in equation **(iii)** we get

*x* = 6.5 – 1.5 *y*

*x* = 6.5 – 1.5(3)

*x* = 6.5 – 4.5

*x* = 2

∴ *x *= 2 and *y* = 3

- Solve 2
*x*+ 3*y*= 11 and 2*x*– 4*y*= – 24 and hence find the value of ‘*m*‘ for which*y*=*mx*+ 3.

**Solution:**

2*x* + 3*y *= 11 … **(i)**

Subtracting 3*y* both side we get

2*x* = 11 – 3*y* … **(ii)**

Putting this value in equation second we get

2*x* – 4*y* = – 24 … **(iii)**

11- 3*y* – 4*y* = – 24

7*y* = – 24 – 11

-7*y* = – 35

*y* = – 35/-7

*y* = 5

Putting this value in equation **(iii)** we get

2*x* = 11 – 3 × 5

2*x* = 11- 15

2*x* = – 4

Dividing by 2 we get

*x* = – 2

Putting the value of *x* and *y*

*y* = *mx* + 3.

5 = -2*m* +3

2*m* = 3 – 5

*m* = -2/2

*m* = -1

**Form the pair of linear equations for the following problems and find their solution by substitution method**

(i) The difference between two numbers is 26 and one number is three times the other. Find them.

**Solution:**

Let larger number = *x*

Smaller number = y

The difference between two numbers is 26

*x* – *y* = 26

*x* = 26 + *y*

Given that one number is three times the other

So *x* = 3*y*

Putting the value of *x* we get

26*y* = 3*y*

-2*y* = – 2 6

*y* = 13

So value of *x* = 3*y*

Putting value of *y*, we get

*x* = 3 × 13 = 39

Hence the numbers are 13 and 39.

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

**Solution:**

Let first angle = *x*

And second number = *y*

As both angles are supplementary so that sum will 180

*x* + *y* = 180

x = 180 – *y* … **(i)**

Difference is 18 degree so that

*x* – *y* = 18

Putting the value of *x *we get

180 – *y* – *y* = 18

– 2*y* = -162

*y* = -162/-2

*y* = 81

Putting the value back in equation **(i)**, we get

*x* = 180 – 81 = 99Hence, the angles are 99º and 81º.

(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.

**Solution:**

Let cost of each bat = Rs *x*

Cost of each ball = Rs *y*

Given that coach of a cricket team buys 7 bats and 6 balls for Rs 3800.

7*x* + 6*y* = 3800

6*y* = 3800 – 7*x*

Dividing by 6, we get

*y* = (3800 – 7*x*)/6 … **(i)**

Given that she buys 3 bats and 5 balls for Rs 1750 later.

3*x* + 5*y *= 1750

Putting the value of *y*

3*x* + 5 ((3800 – 7*x*)/6) = 1750

Multiplying by 6, we get

18*x* + 19000 – 35*x* = 10500

-17*x* =10500 – 19000

-17*x* = -8500

*x* = – 8500/- 17

*x *= 500

Putting this value in equation **(i)** we get

*y *= ( 3800 – 7 × 500)/6

*y *= 300/6

*y *= 50

Hence cost of each bat = Rs 500 and cost of each balls = Rs 50.

(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for traveling a distance of 25 km?

**Solution:**

Let the fixed charge for taxi = Rs *x*

And variable cost per km = Rs *y*

Total cost = fixed charge + variable charge

Given that for a distance of 10 km, the charge paid is Rs 105

*x* + 10*y* = 105 … **(i)**

*x* = 105 – 10*y*

Given that for a journey of 15 km, the charge paid is Rs 155

*x* + 15*y* = 155

Putting the value of *x* we get

105 – 10*y* + 15*y* = 155

5*y* = 155 – 105

5*y* = 50

Dividing by 5, we get

*y* = 50/5 = 10

Putting this value in equation **(i)** we get

*x *= 105 – 10 × 10

*x *= 5

People have to pay for traveling a distance of 25 km

= *x* + 25*y*

= 5 + 25 × 10

= 5 + 250

=255

A person have to pay Rs 255 for 25 Km.

(v) A fraction becomes 9/11, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6 . Find the fraction.

**Solution:**

Let Numerator = *x*

Denominator = *y*

Fraction will = *x*/*y*

A fraction becomes 9/11, if 2 is added to both the numerator and the denominator

(*x* + 2)/*y*+2 = 9/11

By Cross multiplication, we get

11*x* + 22 = 9*y* + 18

Subtracting 22 both side, we get

11*x* = 9*y* – 4

Dividing by 11, we get

*x *= 9*y* – 4/11 … **(i)**

Given that 3 is added to both the numerator and the denominator it becomes 5/6.

If, 3 is added to both the numerator and the denominator it becomes 5/6

(*x*+3)/*y *+3 = 5/6 … **(ii)**

By Cross multiplication, we get

6*x* + 18 = 5*y* + 15

Subtracting the value of *x,* we get

6(9*y* – 4 )/11 + 18 = 5*y* + 15

Subtract 18 both side we get

6(9*y* – 4 )/11 = 5*y* – 3

54 – 24 = 55*y *– 33

–*y* = -9

*y* = 9

Putting this value of y in equation **(i)**, we get

*x *= 9*y* – 4

11 … **(i)**

*x* = (81 – 4)/77

*x* = 77/11

*x* = 7

Hence our fraction is 7/9.

(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

**Solution:**

Let present age of Jacob = *x* year

And present Age of his son is = *y* year

Five years hence,

Age of Jacob will = *x* + 5 year

Age of his son will = *y* + 5year

Given that the age of Jacob will be three times that of his son

*x* + 5 = 3(*y *+ 5)

Adding 5 both side, we get

*x* = 3*y* + 15 – 5

*x* = 3*y* + 10 … **(i)**

Five years ago,

Age of Jacob will = *x* – 5 year

Age of his son will = *y* – 5 year

Jacob’s age was seven times that of his son

*x* – 5 = 7(*y* -5)

Putting the value of *x* from equation **(i)** we get

3*y *+ 10 – 5 = 7*y* – 35

3*y* + 5 = 7*y* – 35

3*y* – 7*y* = -35 – 5

-4y = – 40

*y* = – 40/- 4

*y* = 10 year

Putting the value of *y* in equation first we get

*x* = 3 × 10 + 10

*x* = 40 years

Hence, Present age of Jacob = 40 years and present age of his son = 10 years.

**Exercise 3.4**

**Solve the following pair of linear equations by the elimination method and the substitution method:**

(i)*x*+*y*=5 and 2*x*–3*y*= 4

(ii) 3*x*+ 4*y*= 10 and 2*x*– 2*y*= 2

(iii) 3*x*– 5*y*– 4 = 0 and 9*x*= 2*y*+ 7

(iv)*x*/2 + 2*y*/3 = – 1 and*x*–*y*/3 = 3

**Solution:**

(i) *x* + *y* =5 and 2*x* –3*y* = 4

__By elimination method__

*x* + *y* =5 … **(i)**

2*x* –3*y* = 4 … **(ii)**

Multiplying equation **(i)** by **(ii)**, we get

2*x* + 2*y* = 10 … **(iii)**

2*x* –3*y* = 4 … **(ii)**

Subtracting equation **(ii)** from equation **(iii)**, we get

5*y* = 6

*y* = 6/5

Putting the value in equation **(i)**, we get

*x* = 5 – (6/5) = 19/5

Hence, *x* = 19/5 and *y* = 6/5

__By substitution method__*x* + *y* = 5 … **(i)**

Subtracting *y* both side, we get

*x* = 5 – *y* … **(iv)**

Putting the value of *x* in equation **(ii)** we get

2(5 – *y*) – 3*y* = 4

-5*y* = – 6

*y* = -6/-5 = 6/5

Putting the value of *y* in equation **(iv)** we get

*x* = 5 – 6/5

*x* = 19/5

Hence, *x* = 19/5 and *y* = 6/5 again

(ii) 3*x* + 4*y* = 10 and 2*x* – 2*y* = 2

__By elimination method__

3*x* + 4*y* = 10 …. **(i)**

2*x* – 2*y* = 2 … **(ii)**

Multiplying equation **(ii)** by 2, we get

4*x* – 4*y* = 4 … **(iii)**

3*x* + 4*y* = 10 … **(i)**

Adding equation **(i)** and **(iii)**, we get

7*x* + 0 = 14

Dividing both side by 7, we get

*x* = 14/7 = 2

Putting in equation **(i)**, we get

3*x* + 4*y* = 10

3(2) + 4*y* = 10

6 + 4*y* = 10

4*y* = 10 – 6

4*y* = 4

*y* = 4/4 = 1

Hence, answer is *x* = 2, *y* = 1

__By substitution method__

3*x* + 4*y* = 10 … **(i)**

Subtract 3*x* both side, we get

4*y* = 10 – 3*x*

Divide by 4 we get

*y* = (10 – 3*x* )/4

Putting this value in equation **(ii)**, we get

2*x* – 2*y* = 2 … **(i)**

2*x* – 2(10 – 3*x* )/4) = 2

Multiply by 4 we get

8*x* – 2(10 – 3*x*) = 8

8*x* – 20 + 6*x* = 8

14*x* = 28

*x* = 28/14 = 2

*y *= (10 – 3*x*)/4

*y *= 4/4 = 1

Hence, answer is *x* = 2, *y* = 1 again.

(iii) 3*x* – 5*y* – 4 = 0 and 9*x* = 2*y* + 7

__By elimination method__

3*x* – 5*y* – 4 = 0

3*x* – 5*y* = 4 …**(i)**

9*x* = 2*y* + 7

9*x* – 2*y *= 7 … **(ii)**

Multiplying equation **(i)** by 3, we get

9 x – 15 y = 11 … **(iii)**

9x – 2y = 7 … **(ii)**

Subtracting equation **(ii)** from equation **(iii)**, we get

-13*y* = 5

*y* = -5/13

Putting value in equation **(i)**, we get

3x – 5y = 4 … **(i)**

3x – 5(-5/13) = 4

Multiplying by 13 we get

39*x* + 25 = 52

39*x* = 27

*x* =27/39 = 9/13

Hence our answer is *x* = 9/13 and *y* = – 5/13

__By substitution method__

3*x* – 5*y* = 4 … **(i)**

Adding 5*y* both side we get

3*x* = 4 + 5*y*

Dividing by 3 we get

*x* = (4 + 5*y* )/3 … **(iv)**

Putting this value in equation **(ii)** we get

9*x* – 2*y* = 7 … **(ii)**

9 ((4 + 5*y *)/3) – 2*y* = 7

Solve it we get

3(4 + 5*y* ) – 2*y* = 7

12 + 15*y* – 2*y* = 7

13*y* = – 5

*y* = -5/13

Hence we get *x* = 9/13 and *y* = – 5/13 again.

(iv) *x*/2 + 2*y*/3 = – 1 and *x* – *y*/3 = 3

__By elimination method__

*x*/2 + 2*y*/3 = -1 … **(i)**

*x* – *y*/3 = 3 … **(ii)**

Multiplying equation **(i)** by 2, we get

*x* + 4*y*/3 = – 2 … **(iii)**

*x* – *y*/3 = 3 … (ii)

Subtracting equation **(ii)** from equation **(iii)**, we get

5*y*/3 = -5

Dividing by 5 and multiplying by 3, we get

*y *= -15/5

*y *= – 3

Putting this value in equation **(ii)**, we get

*x* – *y*/3 = 3 … **(ii)**

*x* – (-3)/3 = 3

*x* + 1 = 3

*x* = 2

Hence our answer is *x* = 2 and *y* = −3.

__By substitution method__

*x* – *y*/3 = 3 … **(ii)**

Add *y*/3 both side, we get

*x *= 3 + *y*/3 … **(iv)**

Putting this value in equation** (i)** we get

*x*/2 + 2*y*/3 = – 1 … **(i)**

(3+ *y*/3)/2 + 2*y*/3 = -1

3/2 + *y*/6 + 2*y*/3 = – 1

Multiplying by 6, we get

9 + *y* + 4*y *= – 6

5*y* = -15

*y* = – 3

Hence our answer is *x* = 2 and *y* = −3.

**Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:**

(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 1/2 if we only add 1 to the denominator. What is the fraction?

(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

(iv) Meena went to bank to withdraw Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes of Rs 50 and Rs 100 she received.

(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs 27 for a book kept for seven days, while Susy paid Rs 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

**Solution:**

(i) Let the fraction be *x*/*y*

According to the question,*x* + 1/*y* – 1 = 1

⇒ *x *– *y *= -2 … **(i)***x*/*y*+1 = 1/2

⇒ 2*x* – *y *= 1 … **(ii)**

Subtracting equation **(i)** from equation **(ii)**, we get

*x* = 3 … **(iii)**

Putting this value in equation **(i)**, we get

3 – *y* = -2

–*y* = -5

*y* = 5

Hence, the fraction is 3/5

(ii) Let present age of Nuri = *x*

and present age of Sonu = *y*

According to the given information,question,(*x* – 5) = 3(*y* – 5)

*x* – 3*y* = -10 … **(i)**

(*x* + 10y) = 2(*y* + 10)

*x* – 2*y* = 10 … **(ii)**

Subtracting equation **(i)** from equation **(ii)**, we get

*y* = 20 … **(iii)**

Putting this value in equation **(i)**, we get

*x* – 60 = -10

*x *= 50

Hence, age of Nuri = 50 years and age of Sonu = 20 years.

(iii) Let the unit digit and tens digits of the number be *x *and *y*respectively.

Then, number = 10*y* + *x*

Number after reversing the digits = 10*x* + *y*

According to the question,

x + *y* = 9 … **(i)**

9(10*y *+ *x*) = 2(10*x* + *y*)

88*y* – 11*x* = 0

– *x* + 8*y* =0 … **(ii)**

Adding equation **(i)** and **(ii)**, we get

9*y* = 9

*y* = 1 … **(iii)**

Putting the value in equation **(i)**, we get

*x* = 8

Hence, the number is 10*y* +* x* = 10 × 1 + 8 = 18.

(iv) Let the number of Rs 50 notes and Rs 100 notes be *x* and *y*respectively.

According to the question,

*x *+ *y *= 25 … **(i)**

50*x* + 100*y* = 2000 … **(ii)**

Multiplying equation (i) by 50, we get

50*x* + 50*y* = 1250 … **(iii)**

Subtracting equation **(iii)** from equation **(ii)**, we get

50*y* = 750

*y* = 15

Putting this value in equation **(i)**, we have *x* = 10

Hence, Meena has 10 notes of Rs 50 and 15 notes of Rs 100.

(v) Let the fixed charge for first three days and each day charge thereafter be Rs *x *and Rs *y *respectively.

According to the question,

*x *+ 4*y* = 27 … **(i)**

*x *+ 2*y* = 21 …** (ii)**

Subtracting equation **(ii)** from equation **(i)**, we get

2*y* = 6

*y* = 3 … **(iii)**

Putting in equation **(i)**, we get

*x* + 12 =27

*x* = 15

Hence, fixed charge = Rs 15 and Charge per day = Rs 3.

**Exercise 3.5 **

**Which of the following pairs of linear equations has unique solution, no solution or infinitely many solutions? In case there is a unique solution, find it by using cross multiplication method.**

(i) *x* – 3*y* – 3 = 0 ; 3*x* – 9*y* – 2 =0

(ii) 2*x* + *y* = 5 ; 3*x* +2*y* =8

(iii) 3*x* – 5*y* = 20 ; 6*x* – 10*y* =40

(iv) *x* – 3*y* – 7 = 0 ; 3*x* – 3*y* – 15= 0

**Solution:**

(i) *x* – 3*y* – 3 = 0

3*x* – 9*y* – 2 =0

*a**₁*/*a**₂*= 1/3

*b**₁*/*b**₂*= -3/-9 = 1/3 and

*C**₁*/*C**₂* = -3/-2 = 3/2

*a**₁*/a₂= *b**₁*/*b**₂*≠_{ }*C**₁*/*C**₂*

_{ }Therefore, the given sets of lines are parallel to each other. Therefore, they will not intersect each other and thus, there will not be any solution for these equations.

(ii) 2*x* + *y* = 5

3*x* +2*y* = 8

*a**₁*/*a**₂*= 2/3

*b**₁*/*b**₂*= 1/2 and

*C**₁*/*C**₂* = -5/-8 = 5/8

*a**₁*/a₂≠_{ }*b**₁*/*b*_{2}

_{
}Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations.

By cross-multiplication method,

*x*/*b**₁**C**₂*–*b*_{2}*C**₁*_{ }=_{ }*y*/*C**₁**a*_{2}–*C**₂**a**₁*_{ }= 1/*a**₁**b*_{2}–*a*_{2}*b**₁*

*x*/-8-(-10) = *y*/-15+16 = 1/4-3

*x*/2 = *y*/1 = 1

*x*/2 = 1, *y*/1 = 1

∴ *x *= 2, *y *= 1.

(iii) 3*x* – 5*y* = 20

6*x* – 10*y* = 40

*a**₁*/*a**₂*= 3/6 = 1/2

*b**₁*/*b**₂*= -5/-10 = 1/2 and

*C**₁*/*C**₂* = -20/-40 = 1/2

*a**₁*/a₂=_{ }*b**₁*/*b**₂*=_{ }*C**₁*/*C**₂*

_{
}Therefore, the given sets of lines will be overlapping each other i.e., the lines will be coincident to each other and thus, there are infinite solutions possible for these equations.

(iv) *x* – 3*y* – 7 = 0

3*x* – 3*y* – 15= 0

*a**₁*/*a**₂*= 1/3

*b**₁*/*b**₂*= -3/-3 = 1 and

*C**₁*/*C**₂* = -7/-15 = 7/15

*a**₁*/a₂≠_{ }*b**₁*/*b*_{2}

_{
}Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations.

By cross-multiplication,

*x*/45-(21) = y/-21-(-15) = 1/-3-(-9)

*x*/24 = *y*/-6 = 1/6

*x*/24 = 1/6 and *y*/-6 = 1/6

*x *= 4 and *y *= -1

∴ *x *= 4, *y *= -1.

**(i) For which values of a and b does the following pair of linear equations have an infinite number of solutions?**

2*x*+ 3*y*=7

(*a* – *b*)*x* + (*a* + *b*)*y* = 3*a* +*b* –2

**Solution:**

2*x* + 3*y* -7 = 0

(*a* – *b*)*x* + (*a* + *b*)*y* – (3*a* +*b* –2) = 0

*a**₁*/*a**₂*= 2/*a*–*b* = 1/2

*b**₁*/*b**₂*= -7/*a*+*b* and

*C**₁*/*C**₂* = -7/-(3*a*+*b*-2) = 7/(3*a*+*b*-2)

For infinitely many solutions,*a**₁*/*a**₂*= *b**₁*/*b**₂*_{= }*C**₁*/*C**₂*

_{
}2/*a*–*b *= 7/3*a*+*b*-26*a* + 2*b* – 4 = 7*a* – 7*b*

*a* – 9*b* = -4 … **(i)**

2/*a*–*b *= 3/*a*+*b*

2*a* + 2*b* = 3*a* – 3*b*

*a* – 5*b* = 0 … **(ii)**

Subtracting equation **(i)** from **(ii)**, we get

4*b* = 4

*b* = 1

Putting this value in equation **(ii)**, we get

*a* – 5 × 1 = 0

*a* = 5

Hence, *a* = 5 and *b* = 1 are the values for which the given equations give infinitely many solutions.

(ii) For which value of k will the following pair of linear equations have no solution?

3*x* + *y *= 1

(2*k* –1)*x* + (*k* –1)*y* = 2*k* + 1

**Solution:**

3*x* + *y *-1 = 0

(2*k* –1)*x* + (*k* –1)*y* – (2*k* + 1) = 0

*a**₁*/*a**₂*= 3/2*k*-1

*b**₁*/*b**₂*= 1/*k*-1 and

*C**₁*/*C**₂* = -1/-2*k*-1 = 1/2*k*+1

For no solutions,

*a**₁*/*a**₂*= *b**₁*/*b**₂*≠_{ }*C**₁*/*C**₂*

3/2*k*-1 = 1/*k*-1 ≠ 1/2*k*+1

3/2*k*-1 = 1/*k*-1

3*k* – 3 = 2*k* – 1

*k* = 2

Hence, for *k* = 2, the given equation has no solution.

- Solve the following pair of linear equations by the substitution and cross-multiplication methods:

8*x*+5*y*= 9

3*x*+2*y*= 4

**Solution:**

8*x* +5*y* = 9 … **(i)**

3*x* +2*y* = 4 … **(ii)**

From equation **(ii)**, we get

*x* = 4-2*y*/3 … **(iii)**

Putting this value in equation **(i)**, we get

8(4-2*y*/3) + 5*y* = 9

32 – 16*y* +15*y* = 27

–*y* = -5

*y* = 5 … **(iv)**

Putting this value in equation **(ii)**, we get

3*x* + 10 = 4

*x* = -2

Hence, *x* = -2, *y *= 5

By cross multiplication again, we get

8*x* + 5*y* -9 = 0

3x + 2y – 4 = 0

*x*/-20-(-18) = *y*/-27-(-32) = 1/16-15

*x*/-2 = *y*/5 = 1/1

*x*/-2 = 1 and *y*/5 = 1

*x* = -2 and *y* = 5

**Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method:**

(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay Rs 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs 1180 as hostel charges. Find the fixed charges and the cost of food per day.

**Solution:**Let *x *be the fixed charge of the food and *y *be the charge for food per day.

According to the question,

*x* + 20*y* = 1000 … **(i)**

x + 26*y* = 1180 … **(ii)**

Subtracting equation **(i)** from equation **(ii)**, we get

6*y* = 180

*y* = 180/6 = 30

Putting this value in equation **(i)**, we get

*x* + 20 × 30 = 1000

*x* = 1000 – 600

*x* = 400

Hence, fixed charge = Rs 400 and charge per day = Rs 30

(ii) A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 when 8 is added to its denominator. Find the fraction.

**Solution:**

Let the fraction be *x*/*y*

According to the question,

*x*-1/*y* = 1/3

⇒ 3*x* – *y* = 3… **(i)**

*x*/*y*+8 = 1/4

⇒ 4*x* – *y* = 8 … **(ii)**

Subtracting equation **(i)** from equation** (ii)**, we get

*x* = 5 … **(iii)**

Putting this value in equation **(i)**, we get

15 – *y *= 3

*y *= 12

Hence, the fraction is 5/12.

(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?

**Solution: **Let the number of right answers and wrong answers be *x *and *y *respectively.

According to the question,

3*x* – *y *= 40 … **(i)**

4*x* – 2*y* = 50

⇒ 2*x* – *y *= 25 … **(ii)**

Subtracting equation **(ii)** from equation **(i)**, we get

*x *= 15 … **(iii)**

Putting this value in equation **(ii)**, we get

30 – *y *= 25

Therefore, number of right answers = 15

And number of wrong answers = 5

Total number of questions = 20

And number of wrong answers = 5

Total number of questions = 20

(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?**Solution:**

Let the speed of 1st car and 2nd car be u km/h and v km/h.

Respective speed of both cars while they are travelling in same direction = (*u –* *v*) km/h

Respective speed of both cars while they are travelling in opposite directions i.e., travelling towards each other = (*u *+ *v*) km/h

According to the question,

5(*u *– *v*) = 100

⇒ *u* – *v* = 20 … **(i)**

1(*u* + *v*) = 100 … **(ii)**

Adding both the equations, we get

2*u* = 120

Putting this value in equation **(ii)**, we obtain

*v* = 40 km/h

Hence, speed of one car = 60 km/h and speed of other car = 40 km/h

(v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

**Solution:**

Let length and breadth of rectangle be *x *unit and *y* unit respectively.

Area = *xy*

According to the question,

(*x* – 5) (*y* + 3) = *xy* – 9

⇒ 3*x* – 5*y* – 6 = 0 … **(i)**

(*x* + 3) (*y* + 2) = *xy* + 67

⇒ 2*x* – 3*y* – 61 = 0 … **(ii)**

By cross multiplication, we get

*x*/305-(-18) = *y*/-12-(-183) = 1/9-(-10)

*x*/323 = *y*/171 = 1/19

*x *= 17, *y* = 9

Hence, the length of the rectangle = 17 units and breadth of the rectangle = 9 units.

**Exercise 3.6**

**1. Solve the following pairs of equations by reducing them to a pair of linear equations:**

(i) 1/2*x* + 1/3*y* = 2

1/3*x* + 1/2*y *= 13/6

(ii) 2/√*x* +3/√*y* = 2

4/√*x* – 9/√y = -1

(iii) 4/*x* + 3*y* = 14

3/*x* – 4*y* = 23

(iv) 5/*x*-1 + 1/*y*-2 = 2

6/*x*-1 – 3/*y*-2 = 1

(v) 7*x*-2*y*/*xy* = 5

8*x* + 7*y*/*xy* = 15

(vi) 6*x* + 3*y* = 6*xy*

2*x* + 4*y* = 5*xy*

(vii) 10/*x*+*y* + 2/*x*–*y* = 4

15/*x*+*y* – 5/*x*–*y* = -2

(viii) 1/3*x*+*y* + 1/3*x*–*y* = 3/4

1/2(3*x*–*y*) – 1/2(3*x*–*y*) = -1/8

**Solution:**

(i) 1/2*x* + 1/3*y* = 2

1/3*x* + 1/2*y *= 13/6

Let 1/*x *= *p* and 1/*y *= *q*, then the equations changes as below:

*p*/2 + *q*/3 = 2

⇒ 3*p* + 2*q* -12 = 0 … **(i)**

*p*/3 + *q*/2 = 13/6

⇒ 2*p* + 3*q* -13 = 0 … **(ii)**

By cross-multiplication method, we get

*p*/-26-(-36) = *q*/-24-(-39) = 1/9-4

*p*/10 = *q*/15 = 1/5

*p*/10 = 1/5 and *q*/15 = 1/5

*p *= 2 and *q *= 3

1/*x *= 2 and 1/*y *= 3

Hence, *x *= 1/2 and *y *= 1/3

(ii) 2/√*x* +3/√*y* = 2

4/√*x* – 9/√y = -1

Let 1/√*x *= *p* and 1/√*y* = *q*, then the equations changes as below:

2p + 3q = 2 … **(i)**

4p – 9q = -1 … **(ii)**

Multiplying equation **(i)** by 3, we get

6p + 9q = 6 … **(iii)**

Adding equation **(ii)** and **(iii)**, we get

10p = 5

p = 1/2 … **(iv)**

Putting in equation **(i)**, we get

2 × 1/2 + 3q = 2

3q = 1

q = 1/3*p *= 1/√*x* = 1/2

√*x* = 2

*x* = 4

and

*q *= 1/√*y* = 1/3

√*y* = 3

*y* = 9

Hence,*x* = 4, *y* = 9

3q = 1

q = 1/3

√

and

√

Hence,

(iii) 4/*x* + 3*y* = 14

3/*x* – 4*y* = 23

Putting 1/*x* = *p* in the given equations, we get

4*p* + 3*y* = 14

⇒ 4*p* + 3y – 14 = 0

3*p* – 4*y* = 23

⇒ 3*p* – 4*y* -23 = 0

By cross-multiplication, we get

*p*/-69-56 = *y*/-42-(-92) = 1/-16-9

⇒ –*p*/125 = *y*/50 = -1/25

Now,

–*p*/125 = -1/25 and *y*/50 = -1/25

⇒ *p* = 5 and *y* = -2

Also, *p* = 1/*x* = 5

⇒ *x* = 1/5

So, *x* = 1/5 and *y* = -2 is the solution.

(iv) 5/*x*-1 + 1/*y*-2 = 2

6/*x*-1 – 3/*y*-2 = 1

Putting 1/*x*-1 = *p* and 1/*y*-2 = *q* in the given equations, we obtain

5*p* + *q* = 2 … **(i)**

6*p* – 3*q* = 1 … **(ii)**

Now, by multiplying equation **(i)** by 3 we get

15*p* + 3*q* = 6 … **(iii)**

Now, adding equation **(ii)** and **(iii)**

21*p* = 7

⇒ *p* = 1/3

Putting this value in equation **(ii)** we get,

6×1/3 – 3*q* =1

⇒ 2-3*q* = 1

⇒ -3*q* = 1-2

⇒ -3*q* = -1

⇒ *q* = 1/3

Now,

*p* = 1/*x*-1 = 1/3

⇒1/*x*-1 = 1/3

⇒ 3 = *x *– 1

⇒ *x* = 4

Also,

*q* = 1/*y*-2 = 1/3

⇒ 1/*y*-2 = 1/3

⇒ 3 = *y*-2

⇒ *y* = 5

Hence, *x* = 4 and *y* = 5 is the solution.

(v) 7*x*-2*y*/*xy* = 5

⇒ 7*x*/*xy* – 2*y*/*xy* = 5

⇒ 7/*y* – 2/*x* = 5 … **(i)**

8*x*+7*y*/*xy* = 15

⇒ 8*x*/*xy* + 7*y*/*xy* = 15

⇒ 8/*y* + 7/*x* = 15 … **(ii)**

Putting 1/*x* = *p* and 1/*y* = *q* in **(i)** and **(ii)** we get,

7*q* – 2*p* = 5 … **(iii)**

8*q* + 7*p* = 15 … **(iv)**

Multiplying equation **(iii)** by 7 and multiplying equation **(iv)** by 2 we get,

49*q* – 14*p* = 35 … **(v)**

16*q* + 14*p* = 30 … **(vi)**

Now, adding equation **(v)** and **(vi)** we get,

49*q* – 14*p* + 16*q* + 14*p* = 35 + 30

⇒ 65*q* = 65

⇒ *q* = 1

Putting the value of q in equation **(iv)**

8 + 7*p* = 15

⇒ 7*p* = 7

⇒ *p* = 1

Now,

*p* = 1/*x* = 1

⇒ 1/*x* = 1

⇒ *x* = 1

also, *q* = 1 = 1/*y*

⇒ 1/*y* = 1

⇒ *y *= 1

Hence, *x *=1 and *y* = 1 is the solution.

(vi) 6*x* + 3*y* = 6*xy*

⇒ 6*x*/*xy *+ 3*y*/*xy* = 6

⇒ 6/y + 3/*x* = 6 … **(i)**

2*x* + 4y = 5*xy*

⇒ 2*x*/*xy* + 4y/*xy* = 5

⇒ 2/*y* + 4/*x* = 5 … **(ii)**

Putting 1/*x* = p and 1/*y* = *q* in **(i)** and **(ii)** we get,

6q + 3p – 6 = 0

2q + 4p – 5 = 0

By cross multiplication method, we get

*p*/-30-(-12) = q/-24-(-15) = 1/6-24

*p*/-18 = *q*/-9 = 1/-18

*p*/-18 = 1/-18 and q/-9 = 1/-18

*p *= 1 and q = 1/2

*p *= 1/x = 1 and q = 1/y = 1/2

*x *= 1, y = 2

Hence, *x *= 1 and *y *= 2

(vii) 10/*x*+*y* + 2/*x*–*y* = 4

15/*x*+*y* – 5/*x*–*y* = -2

Putting 1/*x*+*y* = *p* and 1/*x*–*y* = q in the given equations, we get:

10*p* + 2*q* = 4

⇒ 10*p* + 2*q* – 4 = 0 … **(i)**

15*p* – 5*q* = -2

⇒ 15*p* – 5*q* + 2 = 0 … **(ii)**

Using cross multiplication, we get

*p*/4-20 = *q*/-60-(-20) = 1/-50-30

*p*/-16 = *q*/-80 = 1/-80

*p*/-16 = 1/-80 and *q*/-80 = 1/-80

*p* = 1/5 and *q* = 1

*p* = 1/*x*+*y* = 1/5 and *q* = 1/*x*–*y* = 1

*x* + *y* = 5 … **(iii)**

and *x* – *y* = 1 … **(iv)**

Adding equation **(iii)** and **(iv)**, we get

2*x* = 6

*x* = 3 …. **(v)**

Putting value of *x* in equation **(iii)**, we get

*y* = 2

Hence, *x* = 3 and *y* = 2

(viii) 1/3*x*+*y* + 1/3*x*–*y* = 3/4

1/2(3*x*–*y*) – 1/2(3*x*–*y*) = -1/8

Putting 1/3*x*+*y* = *p* and 1/3*x*–*y* = *q* in the given equations, we get

*p* + *q* = 3/4 … **(i)**

*p*/2 – *q*/2 = -1/8

*p *– *q *= -1/4 … **(ii)**

Adding **(i)** and **(ii)**, we get

2*p* = 3/4 – 1/4

2*p* = 1/2

*p *= 1/4

Putting the value in equation **(ii)**, we get

1/4 – *q* = -1/4

*q* = 1/4 + 1/4 = 1/2

*p* = 1/3*x*+*y* = 1/4

3*x* + *y* = 4 … **(iii)**

*q* = 1/3*x*–*y* = 1/2

3*x* – *y* = 2 … **(iv)**

Adding equations **(iii)** and **(iv)**, we get

6*x* = 6

*x* = 1 … **(v)**

Putting the value in equation **(iii)**, we get

3(1) + *y* = 4

*y* = 1

Hence, *x* = 1 and *y* = 1

**2. Formulate the following problems as a pair of equations, and hence find their solutions:**

(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.

**Solution:**

Let the speed of Ritu in still water and the speed of stream be* x*km/h

and *y* km/h respectively.

Speed of Ritu while rowing

Upstream = (*x* – *y*) km/h

Downstream = (*x* + *y*) km/h

According to question,

2(*x* + *y*) = 20

⇒ *x* + *y* = 10 … **(i)**

2(*x* – *y*) = 4

⇒*x* – *y* = 2 … **(ii)**

Adding equation**(i)** and **(ii)**, we get

2(

⇒

Adding equation

Putting this equation in **(i)**, we get

Hence, Ritu’s speed in still water is 6 km/h and the speed of the current is 4 km/h.

(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.

**Solution:**

Let the number of days taken by a woman and a man be *x* and *y*respectively.

Therefore, work done by a woman in 1 day = 1/*x*

According to the question,

4(2/*x* + 5/*y*) = 1

2/*x* + 5/*y* = 1/4

3(3/*x* + 6/*y*) = 1

3/*x* + 6/*y* = 1/3

Putting 1/*x* = *p* and 1/*y *= *q* in these equations, we get

2*p* + 5*q* = 1/4

By cross multiplication, we get

Hence, number of days taken by a woman = 18 and number of days taken by a man = 36

(iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.**Answer**

Let the speed of train and bus be u km/h and v km/h respectively.

According to the given information,

60/*u* + 240/*v *= 4 … **(i)**

100/*u* + 200/*v *= 25/6 … **(ii)**

Putting 1/*u* = *p* and 1/*v* = *q* in the equations, we get

60*p* + 240*q* = 4 … **(iii)**

100*p* + 200*q* = 25/6

600*p* + 1200*q* = 25 … **(iv)**

Multiplying equation **(iii)** by 10, we get

600*p* + 2400*q* = 40 …. **(v)**

Subtracting equation **(iv)** from **(v)**, we get1200*q* = 15

*q* = 15/200 = 1/80 … **(vi)**

Putting equation **(iii)**, we get

60*p* + 3 = 4

60*p* = 1

*p* = 1/60

*p* = 1/*u* = 1/60 and *q* = 1/*v* = 1/80

*u* = 60 and *v* = 80

Hence, speed of train = 60 km/h and speed of bus = 80 km/h.

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]]>- The graphs of y = p(x) are given in following figure, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

**Sol:**

(i) The number of zeroes is 0 as the graph does not cut the *x*-axis at any point.

(ii) The number of zeroes is 1 as the graph intersects the *x*-axis at only 1 point.

(iii) The number of zeroes is 3 as the graph intersects the* x*-axis at 3 points.

(iv) The number of zeroes is 2 as the graph intersects the *x*-axis at 2 points.

(v) The number of zeroes is 4 as the graph intersects the *x*-axis at 4 points.

(vi) The number of zeroes is 3 as the graph intersects the *x*-axis at 3 points.

**Exercise**** 2.2**

- Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i) x²– 2x– 8

(ii) 4s² – 4s + 1

(iii) 6x² – 3 – 7x

(iv) 4u² + 8u

(v) t² – 15

(vi) 3x² – x – 4

**Sol:**

(i) x² – 2x – 8

To make factors given below the method:

The Product of Factors should be: 1 × -8 = -8 (Coefficient of X² × constant value)

And The sum of factors using above value should be -2 (Coefficient of X)

2, 2, 2 (Therefore -4 + 2 = -2, -4 × 2 = -8)

= (x – 4) (x + 2)

The value of x² – 2x – 8 is zero when x – 4 = 0 or x + 2 = 0, i.e., when x = 4 or x = -2

Therefore, the zeroes of x² – 2x – 8 are 4 and -2.

Sum of zeroes = 4 + (-2) = 2 = -(-2)/1 = -(Coefficient of x)/Coefficient of x²

Product of zeroes = 4 × (-2) = -8 = -8/1 = Constant term/Coefficient of x²

(ii) 4s² – 4s + 1

= (2s-1)²^{ }(Using Formula Here: (a-b)² = a² – 2ab + b² )

The value of 4s² – 4s + 1 is zero when 2s – 1 = 0, i.e., s = 1/2

Therefore, the zeroes of 4s² – 4s + 1 are 1/2 and 1/2.

Sum of zeroes = 1/2 + 1/2 = 1 = -(-4)/4 = -(Coefficient ofs)/Coefficient of s²

Product of zeroes = 1/2 × 1/2 = 1/4 = Constant term/Coefficient of s² .

(iii) 6*x²* – 3 – 7*x*

*= *6*x²*^{ }– 7*x *– 3

= 6*x²*^{ }+2*x -9x *– 3

[ i.e 6 × -3 = -18, L.C.M = 2 × 3 × 3, Product of -9 and 2 is -18 & Sum is -7]

= 2*x(3x+1 ) – 3x(3x+1)*

= (3*x* + 1) (2*x* – 3)

The value of 6*x²* – 3 – 7*x* is zero when 3*x* + 1 = 0 or 2*x* – 3 = 0, i.e.,*x* = -1/3 or *x* = 3/2

Therefore, the zeroes of 6*x²* – 3 – 7*x* are -1/3 and 3/2.

Sum of zeroes = -1/3 + 3/2 = 7/6 = -(-7)/6 = -(Coefficient of *x*)/Coefficient of *x²*

Product of zeroes = -1/3 × 3/2 = -1/2 = -3/6 = Constant term/Coefficient of *x²*

(iv) 4*u²* + 8*u*

*= *4*u²* + 8*u *

= 4*u*(*u* + 2)

The value of 4*u²* + 8*u* is zero when 4*u* = 0 or *u* + 2 = 0, i.e., *u* = 0 or *u* = – 2

Therefore, the zeroes of 4*u²* + 8*u* are 0 and – 2.

Sum of zeroes = 0 + (-2) = -2 = -(8)/4 = -(Coefficient of *u*)/Coefficient of *u²*

Product of zeroes = 0 × (-2) = 0 = 0/4 = Constant term/Coefficient of *u²*.

(v) *t²* – 15

= *t²*^{ }– 0.*t* – 15

= (*t *– √15) (*t* + √15)

The value of *t²* – 15 is zero when *t* – √15 = 0 or *t* + √15 = 0, i.e., when *t* = √15 or *t *= -√15

Therefore, the zeroes of *t²* – 15 are √15 and -√15.Sum of zeroes = √15 + -√15 = 0 = -0/1 = -(Coefficient of *t*)/Coefficient of *t*^{2}

Product of zeroes = (√15) (-√15) = -15 = -15/1 = Constant term/Coefficient of *u²*.

(vi) 3*x²* –* x* – 4

= (3*x* – 4) (*x* + 1)

The value of 3*x²* –* x* – 4 is zero when 3*x* – 4 = 0 and *x* + 1 = 0,i.e., when *x* = 4/3 or *x* = -1

Therefore, the zeroes of 3*x²* –* x* – 4 are 4/3 and -1.

Sum of zeroes = 4/3 + (-1) = 1/3 = -(-1)/3 = -(Coefficient of *x*)/Coefficient of *x²*

Product of zeroes = 4/3 × (-1) = -4/3 = Constant term/Coefficient of *x²*.

- Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

(i) 1/4 , -1

(ii) √2 , 1/3

(iii) 0, √5

(iv) 1,1

(v) -1/4 ,1/4

(vi) 4,1

**Sol:**

(i) 1/4 , -1

Let the polynomial be *ax²* + *bx* + *c*, and its zeroes be α and ß

α + ß = 1/4 = –*b*/*a*

αß = -1 = -4/4 = *c*/*a*

If *a* = 4, then *b* = -1, *c* = -4

Therefore, the quadratic polynomial is 4*x²* – *x* -4.

(ii) √2 , 1/3

Let the polynomial be *ax²* + *bx* + *c*, and its zeroes be α and ß

α + ß = √2 = -3√2/3 = –*b*/*a [multiply value by 3/3 to equal value of a with αß ]*

αß = 1/3 = *c*/*a*

If *a* = 3, then *b* = -3√2, *c* = 1

Therefore, the quadratic polynomial is 3*x²* -3√2*x* +1.

or

k { x*² – (sum of zeros)x + Product of zeroes }*

k { x*² – (√2)x + 1/3 }*

k{ 3x*² – 3√2x + 1 }*

(iii) 0, √5

Let the polynomial be *ax²* + *bx* + *c*, and its zeroes be α and ß

α + ß = 0 = 0/1 = –*b*/*a*

αß = √5 = √5/1 = *c*/*a*

If *a* = 1, then *b* = 0, *c* = √5

Therefore, the quadratic polynomial is *x²* + √5.

(iv) 1, 1

Let the polynomial be *ax²* + *bx* + *c*, and its zeroes be α and ß

α + ß = 1 = 1/1 = –*b*/*a*

αß = 1 = 1/1 = *c*/*a*

If *a* = 1, then *b* = -1, *c* = 1

Therefore, the quadratic polynomial is *x²* – *x* +1.

(v) -1/4 ,1/4

Let the polynomial be *ax² *+ *bx* + *c*, and its zeroes be α and ß

α + ß = -1/4 = –*b*/*a*

αß = 1/4 = *c*/*a*

If *a* = 4, then *b* = 1, *c* = 1

Therefore, the quadratic polynomial is 4*x²* + *x* +1.

(vi) 4,1

Let the polynomial be *ax²* + *bx* + *c*, and its zeroes be α and ß

α + ß = 4 = 4/1 = –*b*/*a*

αß = 1 = 1/1 = *c*/*a*

If *a* = 1, then *b* = -4, *c* = 1

Therefore, the quadratic polynomial is *x²* – 4*x* +1.

**Exercise**** 2.3**

- Divide the polynomial
*p*(*x*) by the polynomial*g*(*x*) and find the quotient and remainder in each of the following:

**Sol**

(i) *p*(*x*) = *x³* – 3*x ²* + 5

Quotient = *x*-3 and remainder 7*x* – 9

(ii) *p*(*x*) = *x⁴* – 3*x ²* + 4x + 5,

Quotient = *x ²* +

(iii) *p*(*x*) = *x⁴* – 5*x* + 6, *g*(*x*) = 2 – *x ²*

Quotient = –*x ²* -2 and remainder -5

- Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:

**Sol**

(i) *t ²* – 3, 2

*t ²* – 3 exactly divides 2

(ii) *x ²* + 3

*x ²* + 3

(iii) *x³* – 3*x* + 1, *x⁵* – 4*x³* + *x ²* + 3

*x³* – 3*x* + 1 didn’t divides exactly *x⁵* – 4*x³* + *x ²* + 3

- Obtain all other zeroes of 3
*x*+ 6*⁴**x*– 2*³**x*– 10*²**x*– 5, if two of its zeroes are √(5/3) and – √(5/3).

**Sol:**

*p*(*x*) = 3*x ⁴*+ 6

Since the two zeroes are √(5/3) and – √(5/3).

We factorize *x ²* + 2

Therefore, its zero is given by *x* + 1 = 0 or *x* = -1

As it has the term (*x *+ 1)* ²* , therefore, there will be 2 zeroes at

Hence,the zeroes of the given polynomial are √(5/3), – √(5/3), – 1 and – 1.

- On dividing
*x*– 3*³**x*+*²**x*+ 2 by a polynomial*g*(*x*), the quotient and remainder were*x*– 2 and -2*x*+ 4, respectively. Find*g*(*x*).

**Sol:**

Here in the given question,

Dividend = *x ³*– 3

Quotient = *x* – 2

Remainder = -2*x* + 4

Divisor = *g*(*x*)

We know that,

Dividend = Quotient × Divisor + Remainder

⇒ *x ³*– 3

⇒

⇒

⇒

∴ *g*(*x*) = (*x*^{2} – *x* + 1)

5.Give examples of polynomial *p*(*x*), *g*(*x*), *q*(*x*) and *r*(*x*), which satisfy the division algorithm and

(i) deg *p*(*x*) = deg *q*(*x*)

(ii) deg *q*(*x*) = deg *r*(*x*)

(iii) deg *r*(*x*) = 0

**Sol:**

(i) Let us assume the division of 6*x ² *+ 2

Here,

Degree of

Checking for division algorithm,

Or, 6

Hence, division algorithm is satisfied.

(ii) Let us assume the division of

Here,

Clearly, the degree of

Checking for division algorithm,

Thus, the division algorithm is satisfied.

(iii) Let us assume the division of *x ³*+ 1 by

Here,

g(x) = x

Clearly, the degree of

Checking for division algorithm,

Thus, the division algorithm is satisfied.

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]]>- Use Euclid’s division algorithm to find the HCF of:

(i) 135 and 225

(ii) 196 and 38220

(iii) 867 and 255

**Solution:**

(i) Check Greater Value 225 > 135 ,Always greater value divide with small number

Divide 225 by 135 get 1 quotient and 90 remainder as

225 = 135× 1 + 90

Divide 135 by 90 get 1 quotient and 45 remainder as

135 = 90×1 + 45

Divide 90 by 45 get 2 quotient and NO remainder as

90 = 45 × 2 + 0

As there is no remainder and divisor is 45,so HCF is 45

(ii) 38220 > 196

Divide 38220 by 196 then we get quotient 195 and no remainder

38220 = 196 × 195 + 0

So no remainder, 196 is divisor therefore HCF is 196

(iii) 867 > 255

Divide 867 by 255 we get quotient 3 and remainder 102 ,so

867 = 255 × 3 + 102

Now, divide 255 by 102,then we get quotient is 2 and remainder 51 as

255 = 102 × 2 + 51

Divide 102 by 51, then we get quotient 2 and no remainder,so

102 = 51 × 2 + 0

As there is no remainder ,so divisor is 51 is HCF

2. Show that any positive odd integer is of the form 6q+1, or 6q + 5, where q is some integer.

**Sol:** We know Euclid’s Algorithm is

a = bq + r , Given equation is : 6q + 1,

so b(Divisor) =6, Now we get equation that 6q +r ,r is remainder and q(Quotient) is equal to or more than 0, r = 1,2,3,4,5 because 0 ≤ r ≤ 6

So possible form is

6q + 0, 6 is divisible by 2,so it is a even number

6q+ 1, 6 is divisible by 2 but 1 is not divisible by 2, so it is odd number

Therefore 6q + 5, 6 is divisible by 2 but 5 is not divisible by 2, so it is odd number

3. An army contingent of 616 members is to march behind an army band of 32 members is a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

**Sol:** By HCF (616,32) we can find maximum number of column in which they can march. (HCF means Highest Common Fector)

By Euclid’s Algorithm to find HCF

616 = 32 × 19 + 8

35 = 8 × 4 + 0

The HCF(616,32) is 8.

Therefore, they can march in 8 column each.

4. Use Euclid’s division lemma to show that the square of any integer is either of form 3m or 3m+1 for some integer m.[Hint: Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m + 1.]

**Sol:**

Let a be any positive integer and b = 3.

Then a = 3q + r for some integer q ≥ 0

And r = 0, 1, 2 because 0 ≤ r < 3

Therefore, a = 3q or 3q + 1 or 3q + 2

Or,

a² = (3q)² or (3q + 1)² or (3q + 2)²

a² = 9q² or 9q² + 6q + 1 or 9q² + 12q + 4

= 3 × (3q²) or 3(3q² + 2q) + 1 or 3(3q² + 4q + 1) + 1

= 3k₁ or 3k₂ + 1 or 3k₃ + 1

Where k₁, k₂, k₃ are some positive integers

Hence, it can be said that the square of any positive integer is either of the form 3m or 3m + 1

- Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9
*m*, 9*m*+ 1 or 9*m*+ 8.

**Sol:**

Let a be any positive integer and b = 3

*a* = 3*q* + *r*, where *q* ≥ 0 and 0 ≤ *r* < 3

∴ *a* = 3q or 3*q* + 1 or 3*q* + 2

Therefore, every number can be represented as these three forms. There are three cases.

Case 1: When *a* = 3*q*,

*a*^{3} = (3*q*)^{3} = 27*q*^{3} = 9(3*q*)^{3} = 9*m*,

Where *m* is an integer such that *m* = 3*q*^{3}

Case 2: When *a *= 3q + 1,

*a*^{3} = (3*q* +1)^{3}

*a*^{3}= 27*q*^{3} + 27*q*^{2} + 9*q* + 1

*a*^{3} = 9(3*q*^{3} + 3*q*^{2} + *q*) + 1

*a*^{3} = 9*m* + 1

Where *m* is an integer such that *m* = (3*q*^{3} + 3*q*^{2} + *q*)

Case 3: When *a* = 3*q* + 2,

*a*^{3} = (3*q* +2)^{3}

*a*^{3}= 27*q*^{3} + 54*q*^{2} + 36*q* + 8

*a*^{3} = 9(3*q*^{3} + 6*q*^{2} + 4q) + 8

*a*^{3} = 9*m* + 8

Where *m* is an integer such that *m* = (3*q*^{3} + 6*q*^{2} + 4*q*)

Therefore, the cube of any positive integer is of the form 9*m*, 9*m* + 1,

or 9*m* + 8.

**Excercise 1.2**

- Express each number as product of its prime factors:

(i) 140

(ii) 156

(iii) 3825

(iv) 5005

(v) 7429

**Sol:**

(i) 140 = 2 × 2 × 5 × 7 = 2^{2} × 5 × 7

(ii) 156 = 2 × 2 × 3 × 13 = 2^{2} × 3 × 13

(iii) 3825 = 3 × 3 × 5 × 5 × 17 = 3^{2} × 5^{2} × 17

(iv) 5005 = 5 × 7 × 11 × 13

(v) 7429 = 17 × 19 × 23

- Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.

(i) 26 and 91

(ii) 510 and 92

(iii) 336 and 54

**Sol:**

(i) 26 = 2 × 13

91 =7 × 13

HCF = 13 (HCF is Higher Common Factor. Choose values those exist both sides)

LCM =2 × 7 × 13 =182 (LCF is Least common Factor)

Product of two numbers 26 × 91 = 2366

Product of HCF and LCM 13 × 182 = 2366

Hence, product of two numbers = product of HCF × LCM

Other way to find LCM is:-

LCM :- 13 × 2 × 7 = 182 ( But Our first priority should by divide by 2 then 3 and so on other prime numbers)

(ii) 510 = 2 × 3 × 5 × 17

92 =2 × 2 × 23

HCF = 2

LCM =2 × 2 × 3 × 5 × 17 × 23 = 23460

Product of two numbers 510 × 92 = 46920

Product of HCF and LCM 2 × 23460 = 46920

Hence, product of two numbers = product of HCF × LCM

(iii) 336 = 2 × 2 × 2 × 2 × 3 × 7

54 = 2 × 3 × 3 × 3

HCF = 2 × 3 = 6

LCM = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7 =3024

Product of two numbers 336 × 54 =18144

Product of HCF and LCM 6 × 3024 = 18144

Hence, product of two numbers = product of HCF × LCM.

- Find the LCM and HCF of the following integers by applying the prime factorization method.

(i) 12, 15 and 21

(ii) 17, 23 and 29

(iii) 8, 9 and 25

**Sol:**

(i) 12 = 2 × 2 × 3

15 =3 × 5

21 =3 × 7

HCF = 3

LCM = 2 × 2 × 3 × 5 × 7 = 420

**Excercise 1.3**

- Prove that √5 is irrational.

**Sol:**

Let take √5 as rational number

If *a* and *b* are two co prime number and* b* is not equal to 0.

We can write √5 = *a*/*b*

Multiply by b both side we get

*b*√5 = *a*

To remove root, Squaring on both sides, we get

5*b*^{2} = *a*^{2} … **(i)**

Therefore, 5 divides *a*^{2} (i.e b₂ = a₂ / 5) and according to theorem of rational number, for any prime number *p* which is divides *a*^{2} then it will divide *a* also.

That means 5 will divide *a*. So we can write

*a* = 5*c*

Putting value of *a* in equation **(i)** we get

5*b*^{2} = (5*c*)^{2}

5*b*^{2} = 25*c*^{2}

Divide by 25 we get

*b*2/5 = *c*^{2}

Similarly, we get that *b* will divide by 5

and we have already get that *a* is divide by 5

but *a* and *b* are co prime number. so it contradicts.

Hence √5 is not a rational number, it is irrational.

- Prove that 3 + 2√5 is irrational.

**Sol:**

Let take that 3 + 2√5 is a rational number.

So we can write this number as

3 + 2√5 = *a*/*b*

Here a and b are two co prime number and b is not equal to 0

Subtract 3 both sides we get

2√5 = *a*/*b* – 3

2√5 = (*a*-3*b*)/*b*

Now divide by 2, we get

√5 = (*a*-3*b*)/2*b*

Here *a* and *b* are integer so (*a*-3*b*)/2*b* is a rational number so √5 should be a rational number But √5 is a irrational number so it contradicts.

Hence, 3 + 2√5 is a irrational number.

- Prove that the following are irrationals:

(i) 1/√2 (ii) 7√5 (iii) 6 + √2

**Sol:**

(i) Let take that 1/√2 is a rational number.

So we can write this number as

1/√2 = *a*/*b*

Here *a *and *b* are two co prime number and *b* is not equal to 0

Multiply by √2 both sides we get

1 = (*a*√2)/*b*

Now multiply by *b*

*b* = *a*√2

divide by a we get

*b*/*a* = √2

Here *a* and *b* are integer so b/a is a rational number so √2 should be a rational number But √2 is a irrational number so it contradicts.

Hence, 1/√2 is a irrational number

(ii) Let take that 7√5 is a rational number.

So we can write this number as

7√5 = *a*/*b*

Here *a* and *b* are two co prime number and *b* is not equal to 0

Divide by 7 we get

√5 = *a*/(7*b*)

Here *a* and *b* are integer so *a*/7*b* is a rational number so √5 should be a rational number but √5 is a irrational number so it contradicts. Hence, 7√5 is a irrational number.

(iii) Let take that 6 + √2 is a rational number.

So we can write this number as

6 + √2 = *a*/*b*

Here a and b are two co prime number and b is not equal to 0

Subtract 6 both side we get

√2 = *a*/*b* – 6

√2 = (*a*-6*b*)/*b*

Here *a* and *b* are integer so (a-6*b*)/*b* is a rational number so √2 should be a rational number. But √2 is a irrational number so it contradicts.

Hence, 6 + √2 is a irrational number.

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]]>1. Single Domain Linux Hosting

2. Single Domain Window Hosting

Click to learn about domain name setting

Both are same You can choose any one of above type. I choose “Single Domain Linux Hosting”. Both hosting type can run and use on your Window and Linux system. Actually these are known as share hosting. They provide us single same server for more than one users. Sharing server is less costly.

I suggest purchase Hostgator Hosting for minimum 3 years. Because you will get good discount price. Because renew yearly cost is much higher than first time purchase.

In Single Domain Hosting( Linux or Window) You can tag only one domain name.

After purchase hosting you are not bound to run your blog only with wordpress.org. You have also other software options to create your blog same as WordPress plugin.

**WordPress Hostgator Hosting**

So after purchase you will get mail “Web Hosting Control panel” settings like user name, Password, Name server etc. When you purchase hosting then you have choice to login with your third party domain name or you can buy from same hosting server.

If it is third party domain name then first setting you name server after some time you will able to login your cpanel with domain name like: “example.com/cpanel”

Login your Cpanel.

Click on WordPress.org software and again click on Install Now. After installation login your WordPress.org with your own domain name like:

“example.com/wp-admin”

Remember: After installation remove “WP_” from table Prefix. Leave it blank. Otherwise it will create problem when you publish your first page.

You will able to see your blog with your domain name after creating first post untill showing error with domain name.

I will recommend Hostgator Hosting. Because it have easy tools to use for your blog or website. More important customer care service is very good.

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]]>My aim provides useful and good content to all my readers. That’s why I want to touch all the corners related to blogging. Blogging start is not a big deal. To understand and clear all the doubts of beginners is my provision. Today I am going to clear difference between blog providers’ websites given below:

Most of the Blogger Beginners confuse between WordPress.org and WordPress.com and Blogspot.com. I was one of them.

WordPress.com and Blogger.com are providing free hosting. You can start blogging without any cost using their sub-domain names. Blogger provide easy tools to create new blogs whereas WordPress.com is difficult than blogger to use. Here you can use your own domain name. Hosting is absolutely free which I already discuss to you. But you have not full control on your blog. Blogger can delete your blog any time if they found break any term and condition. The share of AdSense revenue is hidden in these free hosting websites.

WordPress.org is different from above. It is self hosting tools. You have full control on your blog or website. Many people confuse WordPress.com is free and WordPress.org is paid hosting. Both are different from each other. “WordPress.org” is also free to download. “WordPress.org” actually is a software. You need any hosting site to keep your database on web server and use it. WordPress.org software that provide this service to design your blog, add new post, custom design, add new theme design, coding etc and database create on your “cpanel” hosting site.

In above picture you can see that cpanel of Hostgator hosting site provide “WordPress.org” plugin to download. When you will download this software you will be eligible to login on wordpress.org with “www.sitename.com/wpanyfoldername”. Here you can write post, download new plugin, design etc and all database, php code etc store on your hosting provider cpanel.

Are you confused with cpanel?

When you purchase a share hosting from any hosting provider site, they create a cpanel with your domain site name like “www.eptmoney.com/cpanel” and provide username and password.

I hope this post help you clear your all doubts. If you have any confusion than contact me any time at learn@eptmoney.com

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]]>__What is Hosting?__ Most blogger beginners confuse about hosting like me “hahahahhaah”. But seriously I was very confused about hosting until I did not start blogging seriously and not used paid hosing by Hostgator. For example Hosting is like a host of your website or blog. Hosting provides you tools to develop your blog on Internet. Hosting provides space for your blog or website where you can design your own blog. Hosting provides you space where you can store your files, images, data in database online and design your blog, write post, using developing languages like PHP, HTML etc. In hosting you link your domain name address which I already discuss with you.

Are you still confused between Free Hosting and Paid Hosting? What is difference and benefits of it.

Free Hosting: “Blogger” and “WordPress” provides you free hosting tools (Click to know more). You can link here (Blogger or WordPress) your own domain name, write post, modify templates design, modify html code etc. You do not have any need to pay any money for it to use free hosting. Blogger hosting provides easy tools to use and understand for beginners. Any non-technical person can start blogging on blogger to use its tools.

Paid Hosting: You have full control of blog or website by using paid hosting. You have more plugins and tools which you can use for your blog or website. Paid hosting provides many websites like Hostgator, bluehost, godaddy etc. You can also choose domain name from same hosting websites. You pay these companies for hosting and they provide you different tools to design your blog or website. These tools you can use to develop your blog or website. Your all files, data etc stored on provided work area space from where you can add, update or delete these files and edit php,html code. In paid hosting everything provided to you like seo marketing tools (Discuss later about seo), files, Logs, Cloud Flare, Databases, Software’s, App etc.

Paid Hosting Vs Free Hosting : Are you still confusing why we need paid hosting instead of free hosting. There are many reasons to use paid hosting over free hosting.

- In free hosting you have not need to spend any money but in paid hosting you need pay money in use of their services.
- Free hosting have limited templates to design, plugin available for your website or blog whereas paid hosting have many template and also you can download your own new templates, seo tools, many softwares, develop your own code on different platform etc.
- Free hosting website provider can delete your blog any time if they found not follow the term and conditions where as in paid hosting you have full control of blog or website.
- Free hosting provider can change in advertisements showing on your blog, you have not full control of website or blog.
- If you want to be a professional blogger and prepare mind to earn money then you have to use paid hosting rather than free hosting.
- Paid hosting provide many tools and plugin to improve your seo rank to show your blog on first page of google etc whereas its not allowed in free hosting.
- Free hosting is not bad if you just want to just share your thoughts, Information etc or beginner to learn about blogging.
- Blog hosting by Blogger.com, WordPress.com is free whereas to use “WordPress.org” software you have to use paid hosting. Share hosting is less costly for beginners to start blogging with “WordPress.org”

If you can not afford server paid hosting in starting than I will recommend share hosting provided by different hosting provider company like Bluehost, Hostgator etc. In share hosting provided share server more than one user.

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