Find four consecutive integers such that 8 times the sum of the first and the third is 40 greater than 10 times the fourth.

Answer 1

Let our consecutive integers be

n, n+1, n+2, and n+3

Then

1. 8 times the sum of the first and the third means 8(n + n + 2) = 8(2n + 2)
2. 40 greater than 10 times the fourth means 10(n + 3) + 40

So we write

8(2n + 2) = 10(n + 3) + 40 and we solve for n

Expanding the terms with parenthesis on both sides gives

16n + 16 = 10n + 30 + 40 = 10n + 70

Subtracting 16 + 10n from both sides gives

16n – 10n = 70 – 16 which means that 6n = 54

Dividing both sides by 6 gives

n = 9

Answer

Our four consecutive integers are

9, 10, 11, 12

Check

8 times the sum of the first and the third = 8(9 + 11) = 8(20) = 160

40 greater than 10 times the fourth = 10(12) + 40 = 120 + 40 = 160

Answer 2

Let the numbers be x, x+1, x+2, and x+3

8[x + (x+2)] = 10(x + 3) + 40

Solve for x.