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These are equations that require two steps or two operations before arriving at its solutions. In solving two–step equations, there is no definite rule “which” operation to undo first. However, one can apply the rule for order of operations or whichever will result to a simpler equation. Again, the purpose in solving twostep equation is to combine the constants in one side of the equation and see to it that the variable must only have 1 as its numerical coefficient.
Before illustrating the steps on how to solve twostep equations, listed below are examples of two–step equations containing integers.
To solve two–step equations, do the inverse of the operations involved one at a time. Since, we are dealing with equations, it must be noted that whatever operations are done on one side of the equation must as well be done on the other side. Also, we apply the rules for performing operations with integers. To illustrate, consider the following examples:
Solve for the unknown variable:
1.
Solution:
If we divide both sides by 2, we are going to have fractions. It is more preferable to undo subtraction first. Since twice the variable is subtracted by 7 and the opposite of addition is subtraction, then we add 7 to both sides of the equation. Then divide the resulting equation by 2.
2.
Solution:
Again, division of 2 in the equation will lead to fractions. We undo the addition first. Add the opposite of , which is 3 to both sides of the equation then divide both sides by 2.
3.
Solution:
Since the variable term is subtracted by 1, then we add both sides by 1 then divide both sides both sides by 5.
4.
Solution:
Subtract 2 from both sides of the equation then multiply both sides by the inverse of , which is 3.
Practice Exercises:
Solve for the unknown variable.
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To solve onestep equation word problems, follow the following steps:
Below are examples illustrating how to solve onestep equation word problem:
1. When 5 is added to a number, the result is 12. What is the number?
Solution:
Let be the number. The equation is . Solving for a, we get
2. If a number is halved, the result is 24. Find the number.
Solution:
Let be the number. The equation is Solving for , we get
3. Mary has pesos in her wallet. If her mother would give her additional 200 pesos, she will have 230 pesos in all. How much money does she originally have?
Solution:
Since the unknown amount is already as , so the equation is: . Solving for we get
Practice Exercises
1. John bought 10 apples at 12 pesos each. How much do 10 apples cost him?
2. If an item is subject to 20% discount, what is the discount if it was originally marked at 500 pesos?
3. If 10 is subtracted from a number, the difference would be 5. What is the number?
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]]>Still the underlying purpose is to separate the variable and the constant part, by placing the variable on the left side and the constant on the right side of the equation. To do so, perform the opposite of whatever operation is involved in the original equation or by doing the inverse of what is being done to the variable. Since fractions are involved, please note the rules for performing operations with fractions.
Listed below are some examples of one–step equations containing fractions:
Solving One – step Equations Containing Fractions
To solve one–step equation, do the inverse of the operation involved. Since, we are dealing with equations, it must be noted that whatever operation is done on one side of the equation must as well be done on the other side of the equation. Also, apply the rules for performing operations with fractions. To illustrate, consider the following examples:
Solve for the unknown variable:
1.
Solution:
To solve for , we subtract to both sides of the equation.
2.
Solution:
To solve for , add to both sides then simplify the fractions on the right side of the equation.
3.
Solution:
To solve for , multiply both sides by , then simplify the right–hand side of the equation.
4.
Solution:
To solve for , multiply both sides by then simplify the right–hand side.
Practice Exercises:
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Listed below are some examples of one–step equations containing decimals:
To solve one–step equation, do the inverse of the operation involved. Since, we are dealing with equations, it must be noted that whatever operation is done on one side of the equation must as well be done on the other side of the equation. Also, apply the rules for performing operations with decimals. To illustrate, consider the following examples:
Solve for the unknown variable:
1.
Solution:
To solve for , we subtract 0.2 to both sides of the equation.
2.
Solution:
To solve for , add 2.3 to both sides then simplify the decimals on the right side of the equation.
3.
Solution:
To solve for , divide both sides by 3, then simplify the constant part.
4.
Solution:
To solve for , multiply both sides by 0.05 then simplify the constant part.
Practice Exercises
Solve for the unknown variable.
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Solving OneStep Equations Containing Integers
To solve one–step equations, do the inverse of the operation involved. Since, we are dealing with equations, it must be noted that whatever is done on one side of the equation must as well be done on the other side. Also, we apply the rules for performing operations with integers.
To illustrate, consider the following examples:
Solve for the unknown variable:
1.
Solution:
Since the variable is subtracted by a constant 2, we therefore perform the opposite which is to add 2 to both sides of the equation.
2.
Solution:
The variable is added by a negative integer, 3. So we add its opposite, which is +3.
3.
Solution:
Since the variable is multiplied by 2, then we will multiply both sides by its inverse which is .
4.
Solution:
Since is divided by 3, then we will multiply both sides of the equation by 3.
Practice Exercises
Solve for the unknown variable.
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]]>The property says, for any .
Examples:
1.
2.
3.
One of the real conveniences of algebra is the use of parentheses to indicate that certain operations are to be performed. When we do the operations that are indicated, we say "we remove parentheses." For example, the expression indicates that 4 times the binomial is to be subtracted from 3 times the binomial . Hence,
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]]>Using the distributive property, i.e. for all real numbers a, b, and c, a factor multiplying a sum within parentheses should be used to multiply each term of the sum.
Examples:
1.
2.
A sign "+" or "" before an algebraic expression indicates multiplication by or , respectively. Thus, to remove or insert parentheses preceded by a plus sign, we just rewrite the included terms unchanged. If it is preceded by minus sign, we just rewrite the included terms with their signs changed.
Examples:
1.
2.
3.
Two terms with the same literal part are called similar terms regardless of their coefficients, the numerical parts of the terms. If similar terms occur in an algebraic expression and there is no need to apply distributive property, these terms can be added or subtracted by adding or subtracting their numerical coefficients and multiply the result by the common literal part.
Examples:
1.
2.
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]]>An example would be, study the sequence: 2, 4, 8, 16, 32 and so on, what is the sum of the first 12 numbers in this sequence? This problem entails a formula to shorten the procedure. It is on this context, that the core of Algebra lies on representing quantities, patterns or relationships by symbols other than numbers. These symbols, which can be any letter in the English alphabet, are called variables that can take on more than one value. These symbols can further be grouped using the four fundamental operations, which in turn give meaning to equations or inequalities. These groups of symbols are called mathematical expressions which are used to represent verbal expressions or phrases.
Verbal expressions are group of words that comprised a verbal problem or mathematical word problem. Since these would give meaning to the equation/inequality that would represent a problem, one needs to be extra careful in translating any verbal expression into its equivalent mathematical expression.
In this connection, listed below are some of the most commonly used key words to indicate an operation.
Operation  Key Words 

+

plus, add, sum of, increased by, added to, more than, total 


minus, subtract, subtracted from, less, less than, decreased by, reduced by, diminished by, difference between 

times, product, multiply, twice, thrice, doubled, tripled, quadrupled, of, as much as 

divide, ratio, quotient, divided by, divided into, partitioned into 
After learning the key words for each fundamental operation, let’s try to translate some verbal expressions into its equivalent mathematical expression.
Operation  Key Words 

Twice a number  
A number is subtracted from 10  
The sum of 9 and a number  
The quotient of a number and 5  
The ratio of two numbers x and y  
Twice a number decreased by 3  
Twenty percent of a number  
8 more than 3 times a number  
The sum of two numbers divided by 7  
The quotient of twice a number and 5  
The product of a number and 5 decreased by 15  
Thrice a number less than 1  
Half a number  
12 less than thrice a number  
Four times a number more than 20 
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]]>If at given intervals over the whole term of the investment, the principal due is added to the original principal and subsequently grosses interest, the amount by which the original principal has increased by at the conclusion of the term of the investment is called compound interest. The total amount due which comprises the original principal and the compound interest is called the compound amount. Lastly, the time between successive conversions of interest into principal is called conversion period. This implies that under compound interest method, interest earns another interest.
To solve for compound interest, we follow the following steps:
Step 1: Find the future value of the investment or debt using the formula , where F = amount
P = principal
J = the given rate
m = frequency of conversion for 1 year
t = term of investment.
Step 2. Find the interest (I) by using the formula: I = F – P.
Examples:
1. Joey invested P125,540.00 in bank that offers 3.5% compounded semiannually for the first 3 years and 5% compounded quarterly for the next 4 years. If Joey remains his account active in this bank (with no deposits and withdrawals made thereafter), how much is his total money at the end of 7 years? Find his income from this investment.
Solution:
Solving for amount after 3 years
P = P125, 540, j = 3.5%, t = 3, m = 4
Solve for the amount after 4 years using F = P139,374.9412 as P
P = P139,374.9412, j = 5%, t = 4, m = 4
2. Myrna discovered a savings account left to her by her rich foster father while cleaning his old house. When she was 6 years, he invested P50,000 in her name at 5.5% interest compounded annually. Now she is 22 years old. How much money is now available in her account? Find the compound interest.
Solution:
Find the amount in Myrna’s account
Given: P = P50,000, j = 5.5%, m = 1, t = 22 – 6 = 16
Myrna now has P117,763.13 in her account. It means she has earned an interest of P67,763.13.
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]]>To compute for simple interest, multiply the principal, rate and time (expressed in years or a fraction thereof). The sum of the interest and the original principal or amount borrowed is the amount or future value of the investment/obligation.
Consider the following examples on how to compute for interest, amount or principal:
Examples:
1.If P10,000 is invested at 6% then what will $10,000 amount to after 5 years?
Solution:
The future amount (or simply amount) of an investment is the sum of the principal and the interest earned.
Therefore, $10,000 will become $13,000 after 5 years if invested at 6% simple interest.
2. Find the interest at 5% on P5,000 and the corresponding amounts at the end 7 months.
Solution:
Therefore, the interest on $5,000 at 5% for 7 months is $145.83.
3. Mrs. Delos Santos invested P150,000 for two years. For the first year, the rate of interest was 3% and the second year it was 3.5%. How much interest was earned at the end of the two year period?
Solution:
Interest for the first year: Interest = P150,000 (0.03) = P4,500
Interest for the second year: Interest = P150,000(0.035) = P5,250
Therefore, total interest earned from the investment was PP9,750.
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