Math
posted by Kirandeep on .
Find three consecutive integers such that nine less than three times the third is seven more than the sum of the first and second.

ahh...those algebra days...
so, anyways, here's what u do...
1) identify the variables:
Let the 1st integer be "x"
Let the 2nd integer be "(x+1)"
Let the 3rd integer be "(x+2)"
2) now, set up the equation...
3(x+2)  9 = x + (x+1)
3) SOLVE away!!! :D
3x + 6  9 = 2x + 1
3x  3 = 2x + 1
x = 4
4) NOW the three consecutive integers will be...
{ 4, 5, 6, }

YAY, the answer :D
now, DON'T FORGET TO HERP THE DERP!!! 
solve as above, but don't forget the "seven more..."
3(x+2)  9 = x + (x+1) + 7