How do you solve (x(x-4))/(2-3x)<= 3?

Jun 17, 2015

$- 6 \le X \le 1$

Explanation:

Multiplying the numerator by x gives $\frac{{x}^{2} - 4 x}{2 - 3 x} \le 3$
Multiplying by the denominator gives ${x}^{2} - 4 x \le 6 - 9 x$
Rearranging gives ${x}^{2} + 5 x - 6 \le 0$
Factoring gives $\left(x + 6\right) \left(x - 1\right) \le 0$
For the inequality to be true one term must be positive and the other negative
This means $\left(x + 6\right) \le 0 \mathmr{and} \left(x - 1\right) \ge 0$ or $\left(x + 6\right) \ge 0 \mathmr{and} \left(x - 1\right) \le 0$
This means $x \le - 6 \mathmr{and} x \ge 1$ which is impossible
Or $x \ge - 6 \mathmr{and} x \le 1$ which gives $- 6 \le x \le 1$

Jun 17, 2015

Compare to $0$ and do a sign analysis (sign chart, sign table, sign diagram, whatever you were taught to call it).

Explanation:

$\frac{{x}^{2} - 4 x}{2 - 3 x} \le 3$ if and only if:

$\frac{{x}^{2} - 4 x}{2 - 3 x} - 3 \le 0$

Rewrite to get a single ratio on the left.

$\frac{{x}^{2} - 4 x}{2 - 3 x} - \frac{3}{1} \left(\frac{2 - 3 x}{2 - 3 x}\right) \le 0$

$\frac{\left({x}^{2} - 4 x\right) - 3 \left(2 - 3 x\right)}{2 - 3 x} \le 0$

$\frac{\left({x}^{2} - 4 x - 6 + 9 x\right)}{2 - 3 x} \le 0$

$\frac{{x}^{2} + 5 x - 6}{2 - 3 x} \le 0$

Find the key numbers (partition numbers, unnamed special numbers) for the expression on the left. These are the places where the expression might change sign. We find them by finding the zeros and the places where the expression is undefined.

In the end we solve $\text{TOP} = 0$ and $\text{BOTTOM} = 0$

${x}^{2} + 5 x - 6 = \left(x + 6\right) \left(x - 1\right) = 0$ at $x = - 6 , 1$

$2 - 3 x = 0$ at $x = \frac{3}{2}$

The key numbers are:$- 6$, $1$, and $\frac{3}{2}$.

They cut the real number line into intervals:

$\left(- \infty , - 6\right)$, $\left(- 6 , 1\right)$, $\left(1 , \frac{3}{2}\right)$, and $\left(\frac{3}{2} , \infty\right)$

The expression: $\frac{\left(x + 6\right) \left(x - 1\right)}{2 - 3 x}$ is:

positive on $\left(- \infty , - 6\right)$, (test $x = - 10$)
negative on $\left(- 6 , 1\right)$ (test $x = 0$)
positive on $\left(1 , \frac{3}{2}\right)$, (test $x = \frac{5}{4}$)
negative on $\left(\frac{3}{2} , \infty\right)$ (test $x = 5$)

We want the value of $x$ that give negative values for the expression, so the solution is:

$\left(- 6 , 1\right) \cup \left(\frac{3}{2} , \infty\right)$.