EXAMPLE
Solve: [latex]\frac{1}{8}x+\frac{1}{2}=\frac{1}{4}[/latex].
Solution:
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[latex]\frac{1}{8}x+\frac{1}{2}=\frac{1}{4}\quad{LCD=8}[/latex] |
| Multiply both sides of the equation by that LCD, [latex]8[/latex]. This clears the fractions. |
[latex]\color{red}{8(}\frac{1}{8}x+\frac{1}{2}\color{red}{)}=\color{red}{8(}\frac{1}{4}\color{red}{)}[/latex] |
| Use the Distributive Property. |
[latex]8\cdot\frac{1}{8}x+8\cdot\frac{1}{2}=8\cdot\frac{1}{4}[/latex] |
| Simplify — and notice, no more fractions! |
[latex]x+4=2[/latex] |
| Solve using the General Strategy for Solving Linear Equations. |
[latex]x+4\color{red}{-4}=2\color{red}{-4}[/latex] |
| Simplify. |
[latex]x=-2[/latex] |
| Check: Let [latex]x=-2[/latex]
[latex-display]\frac{1}{8}x+\frac{1}{2}=\frac{1}{4}[/latex-display]
[latex-display]\frac{1}{8}(\color{red}{-2})+\frac{1}{2}\stackrel{\text{?}}{=}\frac{1}{4}[/latex-display]
[latex-display]\frac{-2}{8}+\frac{1}{2}\stackrel{\text{?}}{=}\frac{1}{4}[/latex-display]
[latex-display]\frac{-2}{8}+\frac{4}{8}\stackrel{\text{?}}{=}\frac{1}{4}[/latex-display]
[latex-display]\frac{2}{8}\stackrel{\text{?}}{=}\frac{1}{4}[/latex-display]
[latex-display]\frac{1}{4}=\frac{1}{4}\quad\checkmark[/latex-display]
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In the last example, the least common denominator was [latex]8[/latex]. Now it's your turn to find an LCD, and clear the fractions before you solve these linear equations.
Notice that once we cleared the equation of fractions, the equation was like those we solved earlier in this chapter. We changed the problem to one we already knew how to solve!
Example
Solve: [latex]7=\frac{1}{2}x+\frac{3}{4}x-\frac{2}{3}x[/latex].
Answer:
Solution:
We want to clear the fractions by multiplying both sides of the equation by the LCD of all the fractions in the equation.
| Find the least common denominator of all the fractions in the equation. |
[latex]7=\frac{1}{2}x+\frac{3}{4}x-\frac{2}{3}x\quad{LCD=12}[/latex] |
| Multiply both sides of the equation by [latex]12[/latex]. |
[latex]\color{red}{12}(7)=\color{red}{12}\cdot(\frac{1}{2}x+\frac{3}{4}x-\frac{2}{3}x)[/latex] |
| Distribute. |
[latex]12(7)=12\cdot\frac{1}{2}x+12\cdot\frac{3}{4}x-12\cdot\frac{2}{3}x[/latex] |
| Simplify — and notice, no more fractions! |
[latex]84=6x+9x-8x[/latex] |
| Combine like terms. |
[latex]84=7x[/latex] |
| Divide by [latex]7[/latex]. |
[latex]\frac{84}{\color{red}{7}}=\frac{7x}{\color{red}{7}}[/latex] |
| Simplify. |
[latex]12=x[/latex] |
| Check: Let [latex]x=12[/latex]. |
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| [latex]7=\frac{1}{2}x+\frac{3}{4}x-\frac{2}{3}x[/latex]
[latex-display]7\stackrel{\text{?}}{=}\frac{1}{2}(\color{red}{12})+\frac{3}{4}(\color{red}{12})-\frac{2}{3}(\color{red}{12})[/latex-display]
[latex-display]7\stackrel{\text{?}}{=}6+9-8[/latex-display]
[latex-display]7=7\quad\checkmark[/latex-display]
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Now here's a similar problem for you to try. Clear the fractions, simplify, then solve.
Example
Solve: [latex]x+\frac{1}{3}=\frac{1}{6}x-\frac{1}{2}[/latex].
Answer:
Solution:
| Find the LCD of all the fractions in the equation. |
[latex]x+\frac{1}{3}=\frac{1}{6}x-\frac{1}{2},\quad{LCD=6}[/latex] |
| Multiply both sides by the LCD. |
[latex]\color{red}{6}(x+\frac{1}{3})=\color{red}{6}(\frac{1}{6}x-\frac{1}{2})[/latex] |
| Distribute. |
[latex]6\cdot{x}+6\cdot\frac{1}{3}=6\cdot\frac{1}{6}x-6\cdot\frac{1}{2}[/latex] |
| Simplify — no more fractions! |
[latex]6x+2=x-3[/latex] |
| Subtract [latex]x[/latex] from both sides. |
[latex]6x-\color{red}{x}+2=x-\color{red}{x}-3[/latex] |
| Simplify. |
[latex]5x+2=-3[/latex] |
| Subtract 2 from both sides. |
[latex]5x+2\color{red}{-2}=-3\color{red}{-2}[/latex] |
| Simplify. |
[latex]5x=-5[/latex] |
| Divide by [latex]5[/latex]. |
[latex]\frac{5x}{\color{red}{5}}=\frac{-5}{\color{red}{5}}[/latex] |
| Simplify. |
[latex]x=-1[/latex] |
| Check: Substitute [latex]x=-1[/latex]. |
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| [latex]x+\frac{1}{3}=\frac{1}{6}x-\frac{1}{2}[/latex]
[latex-display](\color{red}{-1})+\frac{1}{3}\stackrel{\text{?}}{=}\frac{1}{6}(\color{red}{-1})-\frac{1}{2}[/latex-display]
[latex-display](-1)+\frac{1}{3}\stackrel{\text{?}}{=}-\frac{1}{6}-\frac{1}{2}[/latex-display]
[latex-display]-\frac{3}{3}+\frac{1}{3}\stackrel{\text{?}}{=}-\frac{1}{6}-\frac{3}{6}[/latex-display]
[latex-display]-\frac{2}{3}\stackrel{\text{?}}{=}-\frac{4}{6}[/latex-display]
[latex-display]-\frac{2}{3}=-\frac{2}{3}\quad\checkmark[/latex-display]
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Now you can try solving an equation with fractions that has variables on both sides of the equal sign. The answer may be a fraction.
In the following video we show another example of how to solve an equation that contains fractions and variables on both sides of the equal sign.
https://youtu.be/G5R9jySFMpw
In the next example, we start with an equation where the variable term is locked up in some parentheses and multiplied by a fraction. You can clear the fraction, or if you use the distributive property it will eliminate the fraction. Can you see why?
EXAMPLE
Solve: [latex]1=\frac{1}{2}\left(4x+2\right)[/latex].
Answer:
Solution:
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[latex]1=\frac{1}{2}(4x+2)[/latex] |
| Distribute. |
[latex]1=\frac{1}{2}\cdot4x+\frac{1}{2}\cdot2[/latex] |
| Simplify. Now there are no fractions to clear! |
[latex]1=2x+1[/latex] |
| Subtract 1 from both sides. |
[latex]1\color{red}{-1}=2x+1\color{red}{-1}[/latex] |
| Simplify. |
[latex]0=2x[/latex] |
| Divide by [latex]2[/latex]. |
[latex]\frac{0}{\color{red}{2}}=\frac{2x}{\color{red}{2}}[/latex] |
| Simplify. |
[latex]0=x[/latex] |
| Check: Let [latex]x=0[/latex]. |
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| [latex]1=\frac{1}{2}(4x+2)[/latex]
[latex-display]1\stackrel{\text{?}}{=}\frac{1}{2}(4(\color{red}{0})+2)[/latex-display]
[latex-display]1\stackrel{\text{?}}{=}\frac{1}{2}(2)[/latex-display]
[latex-display]1\stackrel{\text{?}}{=}\frac{2}{2}[/latex-display]
[latex-display]1=1\quad\checkmark[/latex-display]
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Now you can try solving an equation that has the variable term in parentheses that are multiplied by a fraction.
