# How do you graph and solve |4x/3|<2x+5?

Dec 17, 2015

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

#### Explanation:

Since $\left\mid \frac{4}{3} x \right\mid = \frac{4}{3} \left\mid x \right\mid$, we have the following two expressions:

If $x$ is positive:

$\frac{4}{3} x < 2 x + 5$

We can multiply both sides by $3$ to get

$4 x < 6 x + 15$

And finally $2 x > - 15 \setminus \to x > - \frac{15}{2}$

Anyway, we can take the interval $\left(- \frac{15}{2} , \infty\right)$, because we are assuming that $x > 0$. So, the two conditions together yield

$x > 0 \setminus \mathmr{and} x > - \frac{15}{2} \setminus \implies x > 0$

If $x$ isnegative:

$- \frac{4}{3} x < 2 x + 5$

We can multiply both sides by $3$ to get

$- 4 x < 6 x + 15$

And finally $10 x > - 15 \setminus \to x > - \frac{15}{10} = - \frac{3}{2}$

Again, we can't take the whole interval $\left(- \frac{3}{2} , \infty\right)$ because we are assuming that $x < 0$, so

$x < 0 \setminus \mathmr{and} x > - \frac{3}{2} \setminus \implies x \setminus \in \left(- \frac{3}{2} , 0\right)$

So, the inequality is verified from $- \frac{3}{2}$ to $0$, and from $0$ onward.

Moreover, when $x = 0$, we have $0 < 5$, which is true. So, the global solution is $x > - \frac{3}{2}$

Here you can see the graph of the two functions, and as you can see, from $- \frac{3}{2}$ on, $\frac{4}{3} \left\mid x \right\mid$ is below $2 x + 5$.