# How do you solve 5/(x+2)>5/x + 2/3x?

Jun 1, 2017

To work with fractions, our first goal needs to be to find a common denominator.

Let's combine the components on the right of the sign. To do that, we need them to have the same denominator.

$\frac{5}{x} + \frac{2 x}{3}$

Let's multiply $\frac{5}{x}$ by $\frac{3}{3}$ and $\frac{2 x}{3}$ by $\frac{x}{x}$:

$\frac{3}{3} \times \frac{5}{x} + \frac{2 x}{3} \times \frac{x}{x}$

combine

$\frac{3 \times 5}{3 \times x} + \frac{2 \times x \times x}{3 \times x}$

simplify

$\frac{15}{3 x} + \frac{2 {x}^{2}}{3 x}$

combine

$\frac{15 + 2 {x}^{2}}{3 x}$

Now let's put that back into the expression:L

$\frac{5}{x + 2} > \frac{15 + 2 {x}^{2}}{3 x}$

Now, we need to get the same denominator again. Luckily, we already know how to do that!

$\frac{3 x}{3 x} \times \frac{5}{x + 2} > \frac{15 + 2 {x}^{2}}{3 x} \times \frac{x + 2}{x + 2}$

$\frac{3 \times x \times 5}{3 \times x \times \left(x + 2\right)} > \frac{\left(15 + 2 {x}^{2}\right) \times \left(x + 2\right)}{3 \times x \times \left(x + 2\right)}$

simplify

$\frac{15 x}{3 {x}^{2} + 6 x} > \frac{15 x + 30 + 2 {x}^{3} + 4 {x}^{2}}{3 {x}^{2} + 6 x}$

multiply both sides by $\left(3 {x}^{2} + 6 x\right)$

$\cancel{\left(3 {x}^{2} + 6 x\right)} \times \frac{15 x}{\cancel{\left(3 {x}^{2} + 6 x\right)}} > \frac{15 x + 30 + 2 {x}^{3} + 4 {x}^{2}}{\cancel{\left(3 {x}^{2} + 6 x\right)}} \times \cancel{\left(3 {x}^{2} + 6 x\right)}$

$15 x > 2 {x}^{3} + 4 {x}^{2} + 15 x + 30$

Subtract $15 x$ on both sides

$0 > 2 {x}^{3} + 4 {x}^{2} + 30$

factor out a $2$

$0 > 2 \left({x}^{3} + 2 {x}^{2} + 15\right)$

divide by $2$ on both sides

$0 > {x}^{3} + 2 {x}^{2} + 15$