Question #38483

Sep 7, 2016

$x > \frac{39}{11}$

Explanation:

We have: $\frac{2}{3} \left(6 - x\right) < \frac{1}{4} \left(x + 3\right)$

First, let's cross-multiply:

$\implies 8 \left(6 - x\right) < 3 \left(x + 3\right)$

Then, let's expand the parentheses:

$\implies 48 - 8 x < 3 x + 9$

Next, let's subtract $9$ from both sides:

$\implies 39 - 8 x < 3 x$

$\implies 3 x > 39 - 8 x$

Now, let's add $8 x$ to both sides:

$\implies 11 x > 39$

Finally, let's divide both sides by $11$:

$\implies x > \frac{39}{11}$

Therefore, the solution to the inequality is $x > \frac{39}{11}$.