Inequalities Problem : Cancelling out the variables

Question

$\frac{(x+4)}{(x^2-9)} - \frac{2}{(x+3)} < \frac{4x}{(3x-x^2)}$

Below is my solution apprach for this solution :-

$\frac{x+4}{x^2-9} - \frac{2}{x+3} < \frac{4x}{3x-x^2} \\ \Rightarrow \frac{x+4}{x^2-9} - \frac{2}{x+3} < \frac{4}{3-x} \\ \Rightarrow \frac{x+4}{(x+3)(x-3)} - \frac{2}{x+3} < \frac{-4}{x-3} \\ \Rightarrow \frac{x+4}{x+3} - \frac{2(x-3)}{x+3} < -4 \\ \Rightarrow \frac{10-x}{x+3} <-4 \\ \Rightarrow x<-\frac{22}{3} $

But this is not the correct answer. Can someone please tell me what am I missing here?

Answer : $x \in (-\infty,-7.33) \cup (-3,3)-\{0\}$

Answer 1

When you multiply by $x+3$, or $x-3$, it matters whether that is positive or negative. If it is negative, you have to change greater-than to less-than.
Break it into three separate cases: $x\lt-3,-3\lt x\lt3,3\lt x$.

Answer 2

When you are to find all possible values of $x$ meeting the inequality, you should rather expand as is.

$\displaystyle \small \frac{x+4}{x^2-9} - \frac{2}{x+3} \lt \frac{4x}{3x-x^2}$ simplifies to,

$\displaystyle \small \frac{x+4-2x+6}{x^2-9} \lt \frac{4x}{3x-x^2}$

$\displaystyle \small \mathbf {\frac{10-x}{x^2-9} \lt \frac{4}{3-x}} \ $ when $\mathbf {x \ne 0}$

i) When $x \gt 10$, both LHS and RHS are negative. So the inequality becomes

$\displaystyle \small \frac{x-10}{x^2-9} \gt \frac{4}{x-3} \implies 3x \lt -22 $ which is contradictory.

ii) When $3 \lt x \lt 10$, RHS is negative and LHS is positive and hence there are no solutions.

iii) When $-3 \lt x \lt 3$, RHS is positive and LHS is negative so inequality holds.

iv) when $x \lt -3$, both LHS and RHS are positive.

$\displaystyle \small \frac{10-x}{(-x-3)} \lt 4 \implies x \lt -\frac{22}{3} \ \big[ $ note $(-x-3)$ is positive as $\displaystyle \small x \lt -3 \big] $.

Answer 3

You solved as if you were solving an equation for $x$, but from the looks of the solution, the question would have said " find the range of $x$ " $$\frac{(x+4)}{(x^2-9)} - \frac{2}{(x+3)} \lt \frac{4x}{(3x-x^2)}$$ $ x \in \{ -\infty \} $

$$\frac{ (x+4)(x+3)-2(x^2-9)}{(x+3)(x^2-9) } \lt \frac{4x}{(3x-x^2)}$$ $$(3x-x^2) \big( (x+4)(x+3)-2(x^2-9) \big) \lt 4x(x+3)(x^2-9) $$ $$x(3-x)(x+4)(x+3) -2x(3-x)(x+3)(x-3)\lt 4x(x+3)^2(x-3)$$ $$-x(x-3)(x+4)(x+3) +2x(x-3)^2(x+3) \lt 4x(x+3)^2(x-3)$$ $$ (x-3)(x+3) \big[ -x(x+4) +2x(x-3) \lt 4x(x+3) \big] $$ $$ (x-3)(x+3)(\dots \dots )$$ $ x \in \{ 3, -3 \}$ $$ -x^2-4x+2x^2-6x \lt 4x^2+12x $$ $$ 0 \lt 3x^2+22x$$ $$ 0 \lt x(3x+22)$$ $ x \in \{ 0, -\frac{22}{3} \}$