# # Equations with Fractions

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## How to Solve Equations with Fractions

1. Simplify both sides of the equation individually.
2. Multiply both sides of the equation by the least common multiple of all the denominators in the equation.
3. 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 $$\green \frac{3}{4}(9x+\frac{5}{8})=\frac{9}{16}$$.