# How do you solve  6/(x+4 )+ 3/4 = (2x+1) / (3x+12)?

Oct 13, 2015

$x = - 104$

#### Explanation:

Your starting expression looks like this

$\frac{6}{x + 4} + \frac{3}{4} = \frac{2 x + 1}{3 x + 12}$

Notice that you can use $3$ as a common factor for the denominator of the fraction that's on the right-hand side of the equation

$3 x + 12 = 3 \cdot \left(x + 4\right)$

The equation can thus be written as

$\frac{6}{x + 4} + \frac{3}{4} = \frac{2 x + 1}{3 \left(x + 4\right)}$

Next, you need to get rid of the denominators. To do that, you need to find the least common multiple of the expressions that act as denominators for your three fractions

$x + 4 \text{ }$, $\text{ "4" }$, and $\text{ } 3 \left(x + 4\right)$

Notice that you need to multiply the first one by $3$ and by $4$, the second one by $3$ and $\left(x + 4\right)$, and the third one by $4$ to get

$3 \cdot 4 \cdot \left(x + 4\right) \text{ }$

This will be your least common multiple. The equation will become

$\frac{6}{x + 4} \cdot \frac{12}{12} + \frac{3}{4} \cdot \frac{3 \left(x + 4\right)}{3 \left(x + 4\right)} = \frac{2 x + 1}{3 \left(x + 4\right)} \cdot \frac{4}{4}$

$\frac{6 \cdot 12}{12 \left(x + 4\right)} + \frac{3 \cdot 3 \left(x + 4\right)}{12 \left(x + 4\right)} = \frac{\left(2 x + 1\right) \cdot 4}{12 \left(x + 4\right)}$

This is equivalent to

$72 + 9 \cdot \left(x + 4\right) = 4 \cdot \left(2 x + 1\right)$

Finally, expand the parantheses and isolate $x$ on one side of the equation

$72 + 9 x + 36 = 8 x + 4$

$9 x - 8 x = 4 - 108$

$x = \textcolor{g r e e n}{- 104}$

Do a quick check to make sure the calculations are correct

6/(-104 + 4) + 3/4 = (2 * (-104) + 1)/(3 * (-104 + 4)

$- 0.06 + 0.75 = 0.69 \textcolor{w h i t e}{x} \textcolor{g r e e n}{\sqrt{}}$