# Solving equations with fractions

In this section, we will be solving equations such as:       $\frac{x}{4} + \frac{3}{8} = 5$

Let's start by reviewing some basics about fractions.

If we were to multiply the fraction $\frac{3}{8}$ by $8 \$, we would have $8 \cdot \frac{3}{8}$ which is the same as $\frac{8}{1} \times \frac{3}{8}$. Remember, that when fractions are multiplied, we can cancel common factors from top to bottom. It doesn't matter if the factors--numbers that are multiplied--are separated diagonally from top to bottom or are directly above and below each other. Therefore, in this problem, we can cancel the 8s from top to bottom; because 8 goes into 8 once, we write 1s where the 8s were:

$\frac{8}{1} \cdot \frac{3}{8} = \frac{1}{1} \cdot \frac{3}{1}$

Remember that $\frac{1}{1} = 1$, and $\frac{3}{1} = 3$, so

$\frac{1}{1} \cdot \frac{3}{1} = 1 \cdot 3 = 3$

Let's now consider the same expression, but replace the 3 with an x. Recall that the first step is to cancel the 8s from top to bottom and replace with 1s:

$\frac{\color{Red}{8}}{1} \cdot \frac{x}{\color{Red}{8}} =$
$\frac{\color{Red}{1}}{1} \cdot \frac{x}{\color{Red}{1}} =$
$1 \cdot x = x$

Let's use these concepts to simplify the fractions in the equation above.

### Example 1:    $\frac{x}{4} + \frac{3}{8} = 5$

This equation might be simpler to solve if there were no fractional terms. If we were to multiply both sides of the equation (all terms) by something that both 4 and 8 will divide into, we could cancel out common factors from top to bottom, which would eliminate both 4 and 8 in the denominators. The smallest number--least common multiple--that both 4 and 8 will divide into evenly is 8. Let's multiply both sides of the equation by $\frac{8}{1}$.

\begin{align} \frac{x}{4} + \frac{3}{8} &= 5 \\ \frac{\color{Red}{8}}{\color{Red}{1}} \left( \frac{x}{4} + \frac{3}{8} \right) &= \frac{\color{Red}{8}}{\color{Red}{1}} \cdot 5 \qquad \text{multiply both sides by 8/1} \\ \frac{\color{Red}{8}}{\color{Red}{1}} \cdot \frac{x}{4} + \frac{\color{Red}{8}}{\color{Red}{1}} \cdot \frac{3}{8} &= \frac{8}{1} \cdot \frac{5}{1} \qquad \text{distribute 8/1 on the left to both terms in parentheses} \\ \frac{\color{Red}{2}}{1} \cdot \frac{x}{\color{Red}{1}} + \frac{\color{Red}{1}}{1} \cdot \frac{3}{\color{Red}{1}} &= \frac{8}{1} \cdot \frac{5}{1} \qquad \text{8 over 4 reduces to 2 over 1 and 8 over 8 reduces to 1 over 1} \\ 2x + 3 &= 40 \qquad \text{multiply the numerators separately for each term and drop all of the 1s in the denominators} \\ \end{align}

(Note that it is not necessary to write the 1 in the denominator each time we cancel common factors from top to bottom; however, it is sometimes easier to write the 1s in when initially learning to solve fractional equations.)

By multiplying both sides of the equation by the least common factor, we have divided out the denominators. Now we have an equation similar to ones we have previously studied:

\begin{align} 2x + 3 &= 40 \\ 2x + 3 - 3 &= 40 - 3 \qquad \text{to get the term with the variable alone, subtract 3 from both sides} \\ 2x &= 37 \qquad \text{combine like terms} \\ \frac{2x}{2} &= \frac{37}{2} \qquad \text{because x is multiplied by 2, undo the multiplication by dividing both sides by 2} \\ x &= \frac{37}{2} \qquad \text{cancelling the 2s from top to bottom leaves 1 times x which is x} \\ \end{align}

### Example 2:    $\frac{2y}{5} + 3 = \frac{7}{2}$

In this equation, we can eliminate the denominators in the fractional terms by multiplying through the equation by a number that both 5 and 2 will divide into evenly. The smallest such number, the least common multiple, is 10. So, lets multiply each side of the equation by 10. (We will work through this example without putting the 1s in the denominators.)

\begin{align} \frac{2y}{5} + 3 &= \frac{7}{2} \\ {\color{Red}{10}} \cdot \left( \frac{2y}{5} + 3 \right) &= {\color{Red}{10}} \cdot \frac{7}{2} \qquad \text{multiply both sides by 10} \\ \frac{{\color{Red}{10}} \cdot 2y}{5} + {\color{Red}{10}} \cdot 3 &= \frac{{\color{Red}{10}} \cdot 7}{2} \qquad \text{distribute (multiply) 10 on the left to both terms in parentheses} \\ {\color{Red}{2}} \cdot 2y + {\color{Red}{10}} \cdot 3 &= {\color{Red}{5}} \cdot 7 \qquad \text{on the left, 10 over 5 reduces to 2, and on the right, 10 over 2 reduces to 5} \\ 4y + 30 &= 35 \qquad \text{separately multiply each term (PEMDAS--multiplication before addition)} \\ 4y + 30 {\color{Red}- 30} &= 35 {\color{Red}- 30} \qquad \text{to get the term with the variable alone on one side of the equal sign, subtract 30 from both sides} \\ 4y &= 5 \qquad \text{combine like terms} \\ \frac{4y}{\color{Red}{4}} &= \frac{5}{\color{Red}{4}} \qquad \text{divide both sides of the equation by 4} \\ y &= \frac{5}{4} \\ \end{align}

### Example 3:    $\frac{5x}{6} + 5 = \frac{2x}{9} - \frac{3}{2}$

In this equation, we multiply both sides of the equation (all terms) by a number that 6, 9, and 2 will all divide into evenly. The smallest such number, the least common multiple, that all will divide into evenly is 18.

\begin{align} \frac{5x}{6} + 5 &= \frac{2x}{9} - \frac{3}{2} \\ {\color{Red}{18}} \cdot \left( \frac{5x}{6} + 5 \right) &= {\color{Red}{18}} \cdot \left( \frac{2x}{9} - \frac{3}{2} \right) \qquad \text{multiply both sides by 18} \\ \frac{{\color{Red}{18}} \cdot 5x}{6} + {\color{Red}{18}} \cdot 5 &= \frac{{\color{Red}{18}} \cdot 2x}{9} - \frac{{\color{Red}{18}} \cdot 3}{2} \qquad \text{distribute (multiply) 18 to each of the terms inside the parentheses} \\ {\color{Red}{3}} \cdot 5x + {\color{Red}{18}} \cdot 5 &= {\color{Red}{2}} \cdot 2x - {\color{Red}{9}} \cdot 3 \qquad \text{on the left, 18 over 6 reduces to 3, and on the right, 18 over 9 reduces to 2 and 18 over 2 reduces to 9} \\ 15x + 90 &= 4x - 27 \qquad \text{separately multiply each term (PEMDAS--multiplication before addition and subtraction)} \\ 15x {\color{Red}- 4x} + 90 &= 4x {\color{Red}- 4x} - 27 \qquad \text{to get the xs on the same side of the equal sign, subtract 4x from both sides} \\ 11x + 90 &= -27 \qquad \text{combine like terms} \\ 11x + 90 {\color{Red}- 90} &= -27 {\color{Red}- 90}\qquad \text{to get the term with the variable in it alone on one side, subtract 90 from both sides} \\ 11x &= -117 \qquad \text{combine like terms} \\ \frac{11x}{\color{Red}{11}} &= \frac{-117}{\color{Red}{11}} \qquad \text{to get 1x (or x), divide both sides of the equation by 11} \\ 1x &= x = \frac{-117}{11} \\ \end{align}