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.

Math 
herp_derp,
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!!! 
Math 
Steve,
solve as above, but don't forget the "seven more..."
3(x+2)  9 = x + (x+1) + 7