# How do solve (x+2)/(x+5)>=1 algebraically?

Jan 23, 2017

The answer is $x < - 5$

#### Explanation:

Multiply LHS and RHS by ${\left(x + 5\right)}^{2}$

Therefore,

$\frac{x + 2}{x + 5} \cdot {\left(x + 5\right)}^{2} \ge 1 \cdot {\left(x + 5\right)}^{2}$

$\left(x + 2\right) \left(x + 5\right) \ge {\left(x + 5\right)}^{2}$

$\left(x + 2\right) \left(x + 5\right) - {\left(x + 5\right)}^{2} \ge 0$

$\left({x}^{2} + 7 x + 10\right) - \left({x}^{2} + 10 x + 25\right) \ge 0$

$7 x + 10 - 10 x - 25 \ge 0$

$- 3 x - 15 \ge 0$

$3 x \le - 15$

$x \le - 5$

We have to remove the equal sign, as we divide by $0$

So,

$x < - 5$