Find 2 consecutive integer such that 3 times the smaller is 10 more than twice the larger

Question

Struggling with badly with math and need help!

Answer 1

Here's what we know:

b = a + 1 (the second integer is the next number on the number line)

3a = 2b + 10 (3 times the smaller equals 2 times the larger plus another 10)

 

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

3a = 2(a + 1) + 10

distribute

3a = 2a + 2 + 10

simplify:

3a = 2a + 12

subtract 2a:

a = 12

 

You should have no problems figuring out b, and you can check the work to see if it's accurate.