# How do you solve (x+3)/3=(10+4)/4?

Aug 22, 2017

See a solution process below:

#### Explanation:

First, we can simplify the right side of the equation:

$\frac{x + 3}{3} = \frac{14}{4}$

$\frac{x + 3}{3} = \frac{2 \times 7}{2 \times 2}$

$\frac{x + 3}{3} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \times 7}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \times 2}$

$\frac{x + 3}{3} = \frac{7}{2}$

Next, we can multiply each side of the equation by $\textcolor{red}{3}$ to eliminate the fraction on the left side of the equation while keeping the equation balanced:

$\textcolor{red}{3} \frac{x + 3}{3} = \textcolor{red}{3} \times \frac{7}{2}$

$\cancel{\textcolor{red}{3}} \frac{x + 3}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}} = \frac{21}{2}$

$x + 3 = \frac{21}{2}$

Now, we can subtract $\textcolor{red}{3}$ from each side of the equation to solve for $x$ while keeping the equation balanced:

$x + 3 - \textcolor{red}{3} = \frac{21}{2} - \textcolor{red}{3}$

$x + 0 = \frac{21}{2} - \left(\frac{2}{2} \times \textcolor{red}{3}\right)$

$x = \frac{21}{2} - \frac{6}{2}$

$x = \frac{21 - 6}{2}$

$x = \frac{15}{2}$