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

Feb 14, 2017

See the entire solution process below:

#### Explanation:

First, we need to eliminate the fractions by multiplying each side of the equation by a common denominator for the two fractions.

$3 x 4 = \textcolor{red}{12}$ is a common denominator.

$\textcolor{red}{12} \left(\frac{x + 2}{3} + \frac{x - 3}{4}\right) = \textcolor{red}{12} \times 1$

$\left(\textcolor{red}{12} \times \frac{x + 2}{3}\right) + \left(\textcolor{red}{12} \times \frac{x - 3}{4}\right) = 12$

$\left(\cancel{\textcolor{red}{12}} 4 \times \frac{\left(x + 2\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}}\right) + \left(\cancel{\textcolor{red}{12}} 3 \times \frac{\left(x - 3\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}}}\right) = 12$

$4 \left(x + 2\right) + 3 \left(x - 3\right) = 12$

We can now expand the terms in parenthesis and group and combine like terms on the left side of the equation:

$\left(4 \times x\right) + \left(4 \times 2\right) + \left(3 \times x\right) - \left(3 \times 3\right) = 12$

$4 x + 8 + 3 x - 9 = 12$

$4 x + 3 x + 8 - 9 = 12$

$7 x - 1 = 12$

Next, we can add $\textcolor{red}{1}$ to each side of the equation to isolate the $x$ term while keeping the equation balanced:

$7 x - 1 + \textcolor{red}{1} = 12 + \textcolor{red}{1}$

$7 x - 0 = 13$

$7 x = 13$

Now, divide each side of the equation by $\textcolor{red}{7}$ to solve for $x$ while keeping the equation balanced:

$\frac{7 x}{\textcolor{red}{7}} = \frac{13}{\textcolor{red}{7}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{7}}} x}{\cancel{\textcolor{red}{7}}} = \frac{13}{7}$

$x = \frac{13}{7}$