find the intergers

Question

Find two consecutive even integers such that ten times the first equals nine times the second. (Enter your answers as a comma-separated list.)

Answer 1

Here's what we know:

b = a + 2 (the second number is two larger than the first)

10a = 9b (10 times the first is the same as 9 times the second)

 

We have a value for b, so plug it in:

10a = 9(a + 2)

Distribute:

10a = 9a + 18

Subtract 9a from both sides:

10a - 9a = 9a - 9a + 18

a = 18

 

You should be able to find b from here.