Algebra word problems
Example of algebra word problems are numerous. The goal of this unit is to give you the skills that you need to solve a variety of these algebra word problems.A football team lost 5 yards and then gained 9. What is the team's progress?
Solution
For lost, use negative. For gain, use positive.
Progress = -5 + 9 = 4 yards
Example #2:
Use distributive property to solve the problem below:
Maria bought 10 notebooks and 5 pens costing 2 dollars each.How much did Maria pay?
Solution
2 × (10 + 5) = 2 × 10 + 2 × 5 = 20 + 10 = 30 dollars
Example #3:
A customer pays 50 dollars for a coffee maker after a discount of 20 dollars
What is the original price of the coffee maker?
Solution
Let x be the original price.
x - 20 = 50
x - 20 + 20 = 50 + 20
x + 0 = 70
x = 70
Example #4:
Half a number plus 5 is 11.What is the number?
Solution
Let x be the number. Always replace "is" with an equal sign
(1/2)x + 5 = 11
(1/2)x + 5 - 5 = 11 - 5
(1/2)x = 6
2 × (1/2)x = 6 × 2
x = 12
Example #5:
The sum of two consecutive even integers is 26. What are the two numbers?
Solution
Let 2n be the first even integer and let 2n + 2 be the second integer
2n + 2n + 2 = 26
4n + 2 = 26
4n + 2 - 2 = 26 - 2
4n = 24
n = 6
So the first even integer is 2n = 2 × 6 = 12 and the second is 12 + 2 = 14
Below are more complicated algebra word problems
Example #6:
The ratio of two numbers is 5 to 1. The sum is 18. What are the two numbers?
Solution
Let x be the first number. Let y be the second number
x / y = 5 / 1
x + y = 18
Using x / y = 5 / 1, we get x = 5y after doing cross multiplication
Replacing x = 5y into x + y = 18, we get 5y + y = 18
6y = 18
y = 3
x = 5y = 5 × 3 = 15
As you can see, 15/3 = 5, so ratio is correct and 3 + 15 = 18, so the sum is correct.
Example #7: Algebra word problems can be as complicated as example #7. Study it carefully!
Peter has six times as many dimes as quarters in her piggy bank. She has 21 coins in her piggy bank totaling $2.55
How many of each type of coin does she have?
Solution
Let x be the number of quarters. Let 6x be the number of dimes
Since one quarter equals 25 cents, x quarters equals x × 25 cents or 25x cents
Since one dime equals 10 cents, 6x dimes equals 6x × 10 cents or 60x cents
Since one 1 dollar equals 100 cents, 2.55 dollars equals 2.55 × 100 = 255 cents
Putting it all together, 25x cents + 60x cents = 255 cents
85x cents = 255 cents
85x cents / 85 cents = 255 cents / 85 cents
x = 3
6x = 6 × 3 = 18
Therefore Peter has 3 quarters and 18 dimes
Example #8:
The area of a rectangle is x2 + 4x -12. What are the dimensions of the rectangle (length and width)?
Solution
The main idea is to factor x2 + 4x -12
Since -12 = -2 × 6 and -2 + 6 = 4
x2 + 4x -12 = ( x + -2) × ( x + 6)
Since the length is usually longer, length = x + 6 and width = x + -2
Example #9: A must know how when solving algebra word problems
The area of a rectangle is 24 cm2. The width is two less than the length. What is the length and width of the rectangle?
Solution
Let x be the length and let x - 2 be the width
Area = length × width = x × ( x - 2) = 24
x × ( x - 2) = 24
x2 + -2x = 24
x2 + -2x - 24 = 0
Since -24 = 4 × -6 and 4 + -6 = -2, we get:
(x + 4) × ( x + -6) = 0
This leads to two equations to solve:
x + 4 = 0 and x + -6 = 0
x + 4 = 0 gives x = -4. Reject this value since a dimension cannot be negative
x + -6 = 0 gives x = 6
Therefore, length = 6 and width = x - 2 = 6 - 2 = 4
Example #10:
The sum of two numbers is 16. The difference is 4. What are the two numbers?
Let x be the first number. Let y be the second number
x + y = 16
x - y = 4
Solution
Let x be the first number. Let y be the second number
x + y = 16
x - y = 4
Solve the system of equations by elimination
Adding the left sides and the right sides gives:
x + x + y + -y = 16 + 4
2x = 20
x = 10
Since x + y = 16, 10 + y = 16
10 + y = 16
10 - 10 + y = 16 - 10
y = 6
The numbers are 10 and 6
The algebra word problems I solved above are typical questions. You will encounter them a lot in algebra. Hope you had fun solving these algebra word problems.
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