## How to Solve Equations with Fractions

- Simplify both sides of the equation individually.
- Multiply both sides of the equation by the least common multiple of all the denominators in the equation.
- Identify Order of Operations and undo them with inverse operations.

## Examples

Solve for x: \[\frac{2}{3}x-\frac{1}{5}=\frac{7}{10}\]

*Can either side of the equation be simplified?*

\[\blue \frac{2}{3}x-\frac{1}{5}=\frac{7}{10}\]

No, \(\blue \frac{2}{3}x-\frac{1}{5}\) is fully simplified and so is \(\blue \frac{7}{10}\).

*What is the LCM of all the denominators in the equation?*

\[\blue \frac{2}{3}x-\frac{1}{5}=\frac{7}{10}\]

The denominators in the equation are \(\blue 3\), \(\blue 5\), and \(\blue 10\). I can use the list method to find the least common multiple.

\[3, 6, 9, 12, 15, 18, 21, 24, 27, {\blue 30}, 33, 36, …\]

\[5, 10, 15, 20, 25, {\blue 30}, 35, 40, 45, 50, …\]

\[10, 20, {\blue 30}, 40, 50, 60, 70, 80 …\]

The least common multiple of \(\blue 3\), \(\blue 5\), and \(\blue 10\) is \(\blue 30\), so I will multiply both sides of the equation by \(\blue 30\).

\[{\blue 30}(\frac{2}{3}x-\frac{1}{5})={\blue 30}(\frac{7}{10})\]

When I multiply the left side of the equation by \(\blue 30\), I have to distribute it to both of the terms inside the parentheses.

\[{\blue 30}(\frac{2}{3}x)-{\blue 30}(\frac{1}{5})={\blue 30}(\frac{7}{10})\]

Then I can follow the rules for multiplying whole numbers by fractions to simplify each term.

\[\frac{60}{3}x-\frac{30}{5}=\frac{210}{10}\]

When I reduce the fractions, each term should simplify to a whole number, which will make it much easier to solve the equation.

\[20x-6=21\]

*If you knew what \(\blue x\) was, how would you simplify the equation with order of operations?*

\[\blue 20x-6=21\]

If I knew what \(\blue x\) was, I would follow the order of operations like this:

**STEP 1:** Multiply \(\blue x\) by \(\blue 20\).

**STEP 2:** Subtract \(\blue 6\).

To solve the equation, I will undo each step, working backwards.

I will undo **STEP 2** by adding \(\red 6\) to both sides of the equation.

\[\begin{array}{cccc} 20x & -6 & = & 21 \\ & {\red +6} & & {\red +6} \\ \hline 20x & & = & 27 \end{array}\]

Then I will undo **STEP 1** by dividing both sides of the equation by \(\yellow 20\).

\[\begin{array}{ccc} 20x & = & 27 \\ {\yellow \div \, 20} & & {\yellow \div \, 20} \\ \hline x & = & \frac{27}{20} \end{array}\]

\[\blue x=\frac{27}{20}\]

I can check my answer by substituting \(\blue x=\frac{27}{20}\) into the original equation and simplifying to see if it makes a true statement.

\[\frac{2}{3}({\frac{27}{20}})-\frac{1}{5}=\frac{7}{10}\]

Multiply \(\frac{2}{3}\) by \(\frac{27}{20}\).

\[\frac{54}{60}-\frac{1}{5}=\frac{7}{10}\]

Find common denominators for \(\frac{54}{60}\) and \(\frac{1}{5}\).

\[\frac{54}{60}-\frac{12}{60}=\frac{7}{10}\]

Subtract \(\frac{12}{60}\) from \(\frac{54}{60}\).

\[\frac{42}{60}=\frac{7}{10}\]

Reduce \(\frac{42}{60}\).

\[\frac{7}{10}=\frac{7}{10}\]

This is a true statement, so \(\blue x=\frac{27}{20}\) is a solution to \(\blue \frac{2}{3}x-\frac{1}{5}=\frac{7}{10}\).

Solve for x: \[\frac{3}{4}(9x+\frac{5}{8})=\frac{9}{16}\]

*Can either side of the equation be simplified?*

\[\green \frac{3}{4}(9x+\frac{5}{8})=\frac{9}{16}\]

Yes, the left side \(\green \frac{3}{4}(9x+\frac{5}{8})\) can be simplified by distributing the \(\green \frac{3}{4}\).

\[\green \frac{3}{4}(9x+\frac{5}{8})\]

\[\green \frac{27}{4}x+\frac{15}{32}\]

The right side of the equation \(\green \frac{9}{16}\) is fully simplified.

So, the simplified equation is…

\[\green \frac{27}{4}x+\frac{15}{32}=\frac{9}{16}\]

*What is the LCM of all the denominators in the equation?*

\[\green \frac{27}{4}x+\frac{15}{32}=\frac{9}{16}\]

The denominators in the equation are \(\green 4\), \(\green 32\), and \(\green 16\). I can use the list method to find the least common multiple.

\[4, 8, 12, 16, 20, 24, 28, {\green 32}, 36, 40, …\]

\[{\green 32}, 64, 96, 128, 160, …\]

\[16, {\green 32}, 48, 64, 80, 96, …\]

The least common multiple of \(\green 4\), \(\green 32\), and \(\green 16\) is \(\green 32\), so I will multiply both sides of the equation by \(\green 32\).

\[{\green 32}(\frac{27}{4}x+\frac{15}{32})={\green 32}(\frac{9}{16})\]

When I multiply the left side of the equation by \(\green 32\), I have to distribute it to both of the terms inside the parentheses.

\[{\green 32}(\frac{27}{4}x)+{\green 32}(\frac{15}{32})={\green 32}(\frac{9}{16})\]

Then I can follow the rules for multiplying whole numbers by fractions to simplify each term.

\[\frac{864}{4}x+\frac{480}{32}=\frac{288}{16}\]

When I reduce the fractions, each term should simplify to a whole number, which will make it much easier to solve the equation.

\[216x+15=18\]

*If you knew what \(\green x\) was, how would you simplify the equation with order of operations?*

\[\green 216x+15=18\]

If I knew what \(\green x\) was, I would follow the order of operations like this:

**STEP 1:** Multiply \(\green x\) by \(\green 216\).

**STEP 2:** Add \(\green 15\).

To solve the equation, I will undo each step, working backwards.

I will undo **STEP 2** by subtracting \(\red 15\) from both sides of the equation.

\[\begin{array}{cccc} 216x & -15 & = & 18 \\ & {\red -15} & & {\red -15} \\ \hline 216x & & = & 3 \end{array}\]

Then I will undo **STEP 1** by dividing both sides of the equation by \(\yellow 216\).

\[\begin{array}{ccc} 216x & = & 3 \\ {\yellow \div \, 216} & & {\yellow \div \, 216} \\ \hline x & = & \frac{3}{216} \end{array}\]

I can reduce my answer by dividing the numerator and denominator by \(3\).

\[\green x=\frac{1}{72}\]

\[\green x=\frac{1}{72}\]

I can check my answer by substituting \(\green x=\frac{1}{72}\) into the original equation and simplifying to see if it makes a true statement.

\[\frac{3}{4}(9({\green \frac{1}{72}})+\frac{5}{8})=\frac{9}{16}\]

Multiply \(9\) by \(\frac{1}{27}\).

\[\frac{3}{4}(\frac{9}{72}+\frac{5}{8})=\frac{9}{16}\]

Reduce \(\frac{1}{27}\).

\[\frac{3}{4}(\frac{1}{8}+\frac{5}{8})=\frac{9}{16}\]

Add \(\frac{1}{8}\) and \(frac{5}{8}\).

\[\frac{3}{4}(\frac{6}{8})=\frac{9}{16}\]

Multiply \(\frac{3}{4}\) and \(\frac{6}{8}\).

\[\frac{18}{32}=\frac{9}{16}\]

Reduce \(\frac{18}{32}\).

\[\frac{9}{16}=\frac{9}{16}\]

This is a true statement, so \(\green x=\frac{1}{72}\) is a solution to \(\green \frac{3}{4}(9x+\frac{5}{8})=\frac{9}{16}\).