# For what x is 3/(x+3)>(4x)/(x-2)?

Nov 7, 2015

$x \setminus \in \left(- 3 , 2\right)$

#### Explanation:

To simplify the inequality, we are going to perform cross-multiplication of the denominator to avoid dealing with fractions. However, the sign of the inequality might be flipped if we multiply both sides by a negative expression. Hence, we consider $x$ on 3 separate intervals, namely, $\left(- \infty , - 3\right)$, $\left(- 3 , 2\right)$, $\left(2 , \infty\right)$. Note that $x$ cannot -3 or 2. Now, we wish to find the values of $x$ which satisfies

$\frac{3}{x + 3} > \frac{4 x}{x - 2}$

For the case of $x \setminus \in \left(- \infty , - 3\right)$, both $x + 3$ and $x - 2$ are negative.

$3 \left(x - 2\right) > 4 x \left(x + 3\right)$

The sign remains the same as the product of 2 negative expressions is positive.

$3 x - 6 > 4 {x}^{2} + 12 x$

$0 > 4 {x}^{2} + 9 x + 6$

$0 > {x}^{2} + \frac{9}{4} x + \frac{3}{2}$

Completing the square,
$0 > {\left(x + \frac{9}{8}\right)}^{2} + \frac{15}{64}$

Since the RHS is always $\ge \frac{15}{64}$, no real value of $x$ satisfies
$3 \left(x - 2\right) > 4 x \left(x + 3\right)$.

Therefore, no value of $x \setminus \in \left(- \infty , - 3\right)$ satisfies $\frac{3}{x + 3} > \frac{4 x}{x - 2}$.

For the case of $x \setminus \in \left(- 3 , 2\right)$, $x + 3$ is positive while $x - 2$ is negative.

$3 \left(x - 2\right) < 4 x \left(x + 3\right)$

The sign flips as the product of a positive and a negative expressions is negative.

Completing the square,
$0 < {\left(x + \frac{9}{8}\right)}^{2} + \frac{15}{64}$

Since the RHS is always $\ge \frac{15}{64} > 0$, all real value of $x$ satisfies
$3 \left(x - 2\right) > 4 x \left(x + 3\right)$.

Therefore, all values of $x \setminus \in \left(- 3 , 2\right)$ satisfies $\frac{3}{x + 3} > \frac{4 x}{x - 2}$.

For the case of $x \setminus \in \left(2 , \infty\right)$, both $x + 3$ and $x - 2$ are negative.

$3 \left(x - 2\right) > 4 x \left(x + 3\right)$

The sign remains the same as the product of 2 negative expressions is positive.

Completing the square,
$0 > {\left(x + \frac{9}{8}\right)}^{2} + \frac{15}{64}$

Since the RHS is always $\ge \frac{15}{64}$, no real value of $x$ satisfies
$3 \left(x - 2\right) > 4 x \left(x + 3\right)$.

Therefore, no value of $x \setminus \in \left(2 , \infty\right)$ satisfies $\frac{3}{x + 3} > \frac{4 x}{x - 2}$.

Combining the solutions on all 3 intervals, we get

$\frac{3}{x + 3} > \frac{4 x}{x - 2}$

if and only if $x \setminus \in \left(- 3 , 2\right)$.