Verified answer

The previous answer appears to be OK but it is not.

Quick proof; what happens if x = -3.5 ?

LHS = -11, RHS = -5.5 so LHS < RHS

The mistake made is to forget that if you multiply an inequality by something negative then the sign of the inequality is reversed. And of course (x + 3) could be negative so you cannot just multiply by it without taking this into account.

There are two ways to do this problem. One is to divide it into two problems one when (x + 3) is positive and the other when it is negative. The other, slightly faster way is to multiply by (x + 3)^2 which, of course, must be positive. This gives

(2 - x)(x + 3) > (x - 2)(x + 3)^2

(2 - x)(x + 3) - (x - 2)(x + 3)^2 > 0

(2 - x)(x + 3) + (2 - x)(x + 3)^2 > 0

Take out the two common factors to get

(2 - x)(x + 3)(1 + x + 3) > 0

(2 - x)(x + 3)(x + 4) > 0

Now for the LHS to be positive all three factors must be positive which means

-3 < x < 2 or two factors must be negative which happens if x < -4.

The solution is (-inf, -4) U (-3, 2)

Question

(2 – x) / (x + 3) > (x - 2) (What have I done wrong) Help!!?

(x + 3) multiply with (x - 2) on the other side of inequality.

(2 – x) > (x – 2) (x + 3)

2 – x > (x^2 + 3x – 2x – 6)

2 – x > (x^2 + x – 6)

0 > x^2 + x - 6 - 2 + x

0 > x^2 + 2x - 8

0 > (x + 4) (x - 2)

(x + 4) < 0 (or) (x - 2) < 0

x < -4 (or) x < 2

Also, the denominator (x + 3) can not be equal to zero.

x + 3 = 0

x = -3

Thus,

x ≠ - 3

x < -4 (or) x < 2 (and) x ≠ - 3

The answer is:

x < -4 (or) x < 2 (and) x ≠ - 3

-∞ < x < -4 (or) -4 < x < -3 (or) -3 < x < 2

-∞ < x < -3 (or) -3 < x < 2

On line 9, where did the (x + 3) come from? All was OK to there.

Nothing.

There is no x for this inequality.