Find four consecutive integers such that three times the sum of the first two integers exceeds the sum of the last two by 70

Question

The wording is  a bit weird, and I'm not usually very good at consecutive integer word problems. Can anyone show me the steps on how to solve it?

Answer 1

Here's what we know:

We have consecutive integers, so the each number is one more than the previous:

b = a + 1

c = a + 2

d = a + 3

3(a + b) = c + d + 70 (take the sum of the first two integers, multiply it by 3. That's 70 more than the sum of the two larger integers)

 

We have values for our variables, so let's plug them in:

3(a + a + 1) = a + 2 + a + 3 + 70

Simplify:

3(2a + 1) = 2a + 75

Distribute:

6a + 3 = 2a + 75

Time to isolate a. subtract 2a from both sides, and then similarly with 3:

4a + 3 = 75

4a = 72

Divide by 4, and you'll be well on your way to get your answers.