# How do you solve the inequality 3/4+3d < 6 3/4?

Sep 18, 2015

$d < 2$

#### Explanation:

So from $\frac{3}{4} + 3 d < 6 + \frac{3}{4}$ we start by taking the LDC

$\frac{3 + 4 \cdot 3 d}{4} < \frac{6 \cdot 4 + 3}{4}$

Then, we multiply both sides by 4
$4 \cdot \frac{3 + 4 \cdot 3 d}{4} < 4 \cdot \frac{6 \cdot 4 + 3}{4} \rightarrow 3 + 4 \cdot 3 d < 6 \cdot 4 + 3$

The 3s cancel each other out
$3 + 4 \cdot 3 d < 6 \cdot 4 + 3 \rightarrow 4 \cdot 3 d < 6 \cdot 4$

We divide both sides by 4
$\frac{4 \cdot 3 d}{4} < \frac{6 \cdot 4}{4} \rightarrow 3 d < 6$

Finally, divide both sides by 3
$\frac{3 d}{3} < \frac{6}{3} \rightarrow d < 2$