Question

Find three consecutive integers such that when three times the third is increased by the first, the result is 8 less than five times the second.

Answer 1

9, 10 and 11

Explanation:

x - the smallest of three consecutive integers (the first)

x+1 - the next of three consecutive integers (the second)

x+2 - the last of three consecutive integers (the third)

3(x+2) - three times the third integer

3(x+2)+x - three times the third increased by the first

5(x+1) - five times the second tnteger

5(x+1)-8 - 8 less than five times the second

3(x + 2) + x = 5(x + 1) - 8

3x + 6 + x = 5x + 5 - 8

4x + 6 = 5x - 3

-4x -4x

6 = x - 3

x = 9

x+1 = 10

x+2 = 11

Check: 3×11+9 = 33+9 = 42, 5×10-8=50-8 = 42