# 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:

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

JMa beer!