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