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The sum of three numbers is 134. If the third number is 4 times the first

Here is a number problem. If the sum of three numbers is 134, and the third number is 4 times the first, and the first number is 8 less than the second, what are the numbers? Please explain the answer.

Number: 

Assumptions:

x = the second number
x - 8 = the first number
4(x - 8) = third number

Given: Total is 134

first + second + third = 134
x - 8 + x + 4(x - 8) = 134

Simplifying gives:
x - 8 + x + 4x - 32 = 134

Adding like terms:
6x - 40  = 134
6x = 134 + 40
6x = 174
x = 29 => second number

First number =>  x - 8
29 - 8 = 21

Third number => 4(x - 8)
4(29 - 8) = 4(21) = 84

Therefore:
First number => 21
Second number => 29
Third number => 84
 
TOTAL = 21 + 29 + 84 = 134




Assumptions:x = the second

Assumptions:

x = the second number
x - 8 = the first number
4(x - 8) = third number

Given: Total is 134

first + second + third = 134
x - 8 + x + 4(x - 8) = 134

Simplifying gives:
x - 8 + x + 4x - 32 = 134

Adding like terms:
6x - 40  = 134
6x = 134 + 40
6x = 174
x = 29 => second number

First number =>  x - 8
29 - 8 = 21

Third number => 4(x - 8)
4(29 - 8) = 4(21) = 84

Therefore:
First number => 21
Second number => 29
Third number => 84
 
TOTAL = 21 + 29 + 84 = 134