# How do you solve (x+3)/(3x-6)<=2?

May 9, 2018

$x \in \left(- \infty , 2\right) \cup \left[3 , \infty\right)$

#### Explanation:

As per the question, we have

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

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

$\therefore \frac{x + 3 - 6 x + 12}{3 x - 6} \le 0$

$\therefore \frac{- 5 x + 15}{3 x - 6} \le 0$

$\therefore \frac{5 x - 15}{3 x - 6} \ge 0$

$\therefore \frac{5 \left(x - 3\right)}{3 \left(x - 2\right)} \ge 0$

$\therefore$ By Wavy Curve Method we get, Note : In the image the orange region represents the area on the numberline to which $x$ belongs, i.e from $- \infty$ to $2$ and $3$ to $\infty$.

$x \in \left(- \infty , 2\right) \cup \left[3 , \infty\right)$

May 9, 2018

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#### Explanation:

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

Multiply by ${\left(3 x - 6\right)}^{2}$ ( valid since this is always positive )

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

$3 {x}^{2} + 3 x - 18 \le 18 {x}^{2} - 72 x + 72$

$15 {x}^{2} - 75 x + 90 \ge 0$

Factor:

$\left(15 x - 30\right) \left(x - 3\right) \ge 0$

$\left(15 x - 30\right) \left(x - 3\right) = 0 \implies x = 2 , x = 3$

Using signs of brackets:

For:

$2 \le x \le 3$

$+ - \ge 0 \textcolor{w h i t e}{888}$ False

For:

$x \le 2$

$- - \ge 0 \textcolor{w h i t e}{888}$ True

For:

$x \ge 3$

$+ + \ge 0 \textcolor{w h i t e}{888}$ True

Solutions in interval notation:

$\left(- \infty , 2\right) \cup \left[3 , \infty\right)$

Notice the use of a open interval for 2, this is because for $x = 2$ the denominator would be zero.