Answer the word problem with a equation

Question

Find four consecutive integers such that three times the sum of the first two integers

exceeds the sum of the last two by 70.

Answer 1

Let's say the first integer is Y. That means the integers will be:

Y, Y+1, Y+2, and Y+3 (Since we are told that they are consecutive)

Three times the sum of the first 2 integers is:

3(Y+Y+1)

The sum of the last 2 integers is:

Y+2 + Y+3

We are told that 3 times the sum of the first 2 integers exceeds the sum of the last 2 integers by 70, represented as follows:

3(Y+Y+1) - (Y+2 + Y+3) = 70

simplify this equation to solve for Y:

3(2Y+1) - (2Y+5) = 70

6Y+3-2Y-5 = 70

4Y-2 = 70

4Y = 72

Y = 18

So the integers as we defined earlier are: Y, Y+1, Y+2, and Y+3

substituting the value of Y =18

The integers are 18, 19, 20 and 21

We can check to see if we are right by substituting the values into the condition as follows:

3 (18+19) - (20+21)

3(37) - 41

111 - 41 = 70