find three consecutive odd integers such that five times the third is seven less than half the sum of the first and second

Question

I can't figure this word problem out...

Answer 1

x, x+2, and x+4 are the integers

5(x+4)=(0.5)(x+x+2)-7

5x+20=0.5(2x+2)-7

5x+20=x+1-7

5x+20=x-6

5x-x=-20-6

4x=-26 will not yield integers

is there a typo ?

if you use x+4 as the first, x+2 as the second, and x as the third, then you can get integer answers

5x=(1/2)(x+4+x+2)-7

5x=(1/2)(2x+6)-7

5x=x+3-7

5x-x=-4

4x=-4

x=-1, x+2=1, and x+4=3

the integers will be 3, 1, -1