There are 3 consecutive even integers such that the sum of twice the 1st integer and the 3rd integer is ten less than the 2nd integer. What are the 3 even integers?

Answer 1

Three even integers:

n, n + 2, n + 4

2n + n + 4 = n + 2 - 10

Can you solve for n and answer?

Answer 2

2(x+x+4) = x+2 -10 twice sum of 1st and 3rd = 10 less than the 2nd

1st even integer = x, 2nd = x+2, 3rd = x+4

2(2x+4) = 4x+4 = LS, RS = x-8

3x = -4

x = -4/3 no solution if that's the problem It's ambiguous but it must instead be

2x +x+4 = x +2 -10

2x = -4

x = -2, x+2 = 0, x+4 = 2

the consecutive even integers are -2, 0, 2