Question #7a137

2 Answers
Jan 23, 2018

Answer:

The three numbers are #54#, #324# and #154#.

Explanation:

Call the first number '#n#', then the second number is #6n# and the third is #n+100#.

We know the total of the three numbers is #n+6n+n+100=532#.

Collecting like terms, #8n+100=532#, then #8n=432#, so #n=432/8=54#.

The second number is #6# times #n#, so #6xx54=324#.

The third is #n+100=154#.

Jan 23, 2018

Answer:

54, 154, & 324

Explanation:

To solve this, we'll call the first number #x# .
Using this variable, we can create equations for the remaining two equations.
The second number we'll call #y#. It is 6 times the first so
#y=6x# .
The third number we'll call #z#. It is 100 more than the first so
#z=100 + x#.
The sum of all three numbers is #x+y+z=532#.
Substituting in our equations we get #x + (6x) + (100+x) = 532# which simplifies to #8x=432#.
We get #x=54#, #y=6(54)=324#, and #z=100+54=154#.