Posted by **Kirandeep** on Wednesday, October 2, 2013 at 5:22pm.

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**, Wednesday, October 2, 2013 at 6:31pm
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**, Wednesday, October 2, 2013 at 7:27pm
solve as above, but don't forget the "seven more..."

3(x+2) - 9 = x + (x+1) + 7

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