Find three consecutive odd integers such that three times the smallest is 70 more than the largest. What is the sum of the numbers?

Answer 1

smallest odd number is x

next consecutive odd number is x + 2

largest is x + 2 + 2 = x + 4

We also know that 3x = x + 4 + 70

Solving for x: 2x = 74; x = 37

other numbers are x + 2 = 39 and x + 4 = 41.

So sum is 37 + 39 + 41 =117

3(37) = 37 + 4 + 70

Answer 2

The first step is to decide what you want the variable to stand for. I suggest you let x = the smallest number.

The next thing is to represent the second number using x. But be careful, because all of your numbers are odd. If x is the first number, then x + 2 is the second number. [This often confuses people because 2 is not an odd number. But try some real numbers and you see each odd number is 2 more than the one before it.]

The third number = x + 4.

Now set up your equation for "three times the smallest is 70 more than the largest."

3(x) = 70 + (x + 4)

Solve that equation for x. That will tell you what the smallest number is.

[If you have trouble solving that equation, just ask. But try it yourself first.]

Last, they ask for the sum of the three numbers. So, figure out the smallest number (x) and the second number (x+2) and the third number (x+4), then add them together to find the sum.