# Basic inequality problem

• Apr 19th 2009, 01:35 PM
kappa
Linear inequality problem(solved)
Hello, I'm a new member and if I unknowingly made a mistake by posting in the wrong forum, I apologize. I've stumbled upon a problem while studying for a important test and I'd really like to understand it.

I'm supposed to find out what x could be.

(x-3)+x≥3(3-x)

I got this:

(x-3)+x≥3(3-x)
2x-3≥9-3x
2x≥12-3x
-1x≥12
x≥-12

But the answer book says it's x≥2,4
I think there's something wrong with my simplifying.
• Apr 19th 2009, 02:10 PM
stapel
Quote:

Originally Posted by kappa
I'm supposed to find out what x could be.

(x-3)+x≥3(3-x)

The first step for solving this linear inequality will be to multiply things out and see what we've got:

. . . . .(x - 3) + x > 3(3 - x)

. . . . .x - 3 + x > 9 - 3x

. . . . .2x - 3 > 9 - 3x

Let's move everything over to one side:

. . . . .5x - 12 > 0

This is linear, so:

. . . . .5x > 12

. . . . .x > 12/5 = 2.4

:D
• Apr 19th 2009, 02:18 PM
kappa
Quote:

Originally Posted by stapel
The first step for solving this linear inequality will be to multiply things out and see what we've got:

. . . . .(x - 3) + x > 3(3 - x)

. . . . .x - 3 + x > 9 - 3x

. . . . .2x - 3 > 9 - 3x

Let's move everything over to one side:

. . . . .5x - 12 > 0

This is linear, so:

. . . . .5x > 12

. . . . .x > 12/5 = 2.4

:D