Question

Find three consecutive integers such that the sum of the first and twice the second is 80 minus three times the third

Answer 1

Answer:

The three consecutive integers are 12, 13, 14.

Step-by-step explanation:

Encode the question in an equation. We are given two facts: (1) the three integers are consecutive, and (2) they relate to each other through a sum.

Pick any of the integers (first, second, or third) and call it n. Then write the two others correspondingly in terms of n. I pick the middle one as n, so the three consecutive integers are (n-1), n, (n+1). That's (1). As for (2), here is the fact:

$(n-1)+2n = 80-3(n+1)$

Lo and behold, one equation with one unknown! Easy to solve for n to give us  the middle integer, n:

$(n-1)+2n = 80-3(n+1)\\3n-1= 80-3n -3\\6n =78\\n = 13$

The three consecutive integers are 12, 13, 14.