Example
Determine whether each of the following is a solution of [latex]x-\Large\frac{3}{10}=\Large\frac{1}{2}[/latex]
- [latex]x=1[/latex]
- [latex]x=\Large\frac{4}{5}[/latex]
- [latex]x=-\Large\frac{4}{5}[/latex]
Solution:
| 1. |
|
|
[latex]x -\Large\frac{3}{10} =\Large\frac{1}{2}[/latex] |
| Substitute [latex]\color{red}{1}[/latex] for x. |
[latex]\color{red}{1} -\Large\frac{3}{10} =\Large\frac{1}{2}[/latex] |
| Change to fractions with a LCD of [latex]10[/latex]. |
[latex]\color{red}{\Large\frac{10}{10}} -\Large\frac{3}{10} =\Large\frac{5}{10}[/latex] |
| Subtract. |
[latex]\Large\frac{7}{10} \not=\Large\frac{5}{10}[/latex] |
Since [latex]x=1[/latex] does not result in a true equation, [latex]1[/latex] is not a solution to the equation.
| 2. |
|
|
[latex]x -\Large\frac{3}{10} =\Large\frac{1}{2}[/latex] |
| Substitute [latex]\color{red}{\Large\frac{4}{5}}[/latex] for x. |
[latex]\color{red}{\Large\frac{4}{5}} -\Large\frac{3}{10} =\Large\frac{1}{2}[/latex] |
|
[latex]\color{red}{\Large\frac{8}{10}} -\Large\frac{3}{10} =\Large\frac{5}{10}[/latex] |
| Subtract. |
[latex]\Large\frac{5}{10} =\Large\frac{5}{10}\quad\checkmark[/latex] |
Since [latex]x=\Large\frac{4}{5}[/latex] results in a true equation, [latex]\Large\frac{4}{5}[/latex] is a solution to the equation [latex]x-\Large\frac{3}{10}=\Large\frac{1}{2}[/latex].
| 3. |
|
|
[latex]x -\Large\frac{3}{10} =\Large\frac{1}{2}[/latex] |
| Substitute [latex]\color{red}{-\Large\frac{4}{5}}[/latex] for x. |
[latex]\color{red}{-\Large\frac{4}{5}} -\Large\frac{3}{10} =\Large\frac{1}{2}[/latex] |
|
[latex]\color{red}{-\Large\frac{8}{10}} -\Large\frac{3}{10} =\Large\frac{5}{10}[/latex] |
| Subtract. |
[latex]-\Large\frac{11}{10}\not=\Large\frac{5}{10}[/latex] |
Since [latex]x=-\Large\frac{4}{5}[/latex] does not result in a true equation, [latex]-\Large\frac{4}{5}[/latex] is not a solution to the equation.
Addition, Subtraction, and Division Properties of Equality
For any numbers [latex]a,b,\text{ and }c[/latex],
| [latex]\text{if }a=b,\text{ then }a+c=b+c[/latex]. |
Addition Property of Equality |
| [latex]\text{if }a=b,\text{ then }a-c=b-c[/latex]. |
Subtraction Property of Equality |
| [latex]\text{if }a=b,\text{ then }\Large\frac{a}{c}=\Large\frac{b}{c}\normalsize ,c\ne 0[/latex]. |
Division Property of Equality |
In other words, when you add or subtract the same quantity from both sides of an equation, or divide both sides by the same quantity, you still have equality.
Example
Solve: [latex]y+\Large\frac{9}{16}=\Large\frac{5}{16}[/latex]
Answer:
Solution:
|
[latex]y +\Large\frac{9}{16} =\Large\frac{5}{16}[/latex] |
| Subtract [latex]\Large\frac{9}{16}[/latex] from each side to undo the addition. |
[latex]y +\Large\frac{9}{16}\color{red}{-}\color{red}{\Large\frac{9}{16}} =\Large\frac{5}{16}\color{red}{-}\color{red}{\Large\frac{9}{16}}[/latex] |
| Simplify on each side of the equation. |
[latex]y + 0 = -\Large\frac{4}{16}[/latex] |
| Simplify the fraction. |
[latex]y = -\Large\frac{1}{4}[/latex] |
| Check: |
[latex]y +\Large\frac{9}{16} =\Large\frac{5}{16}[/latex] |
|
| Substitute [latex]y=-\Large\frac{1}{4}[/latex] . |
[latex]\color{red}{-\Large\frac{1}{4}} +\Large\frac{9}{16} \stackrel{?}{=}\Large\frac{5}{16}[/latex] |
|
| Rewrite as fractions with the LCD. |
[latex]\color{red}{-\Large\frac{4}{16}} +\Large\frac{9}{16}\stackrel{?}{=}\Large\frac{5}{16}[/latex] |
|
| Add. |
[latex]\Large\frac{5}{16} =\Large\frac{5}{16}\quad\checkmark[/latex] |
|
Since [latex]y=-\Large\frac{1}{4}[/latex] makes [latex]y+\Large\frac{9}{16}=\Large\frac{5}{16}[/latex] a true statement, we know we have found the solution to this equation.
In our next example, we will solve an equation with fractions whose denominators are different. We will need to make an additional step to find the common denominator.
Example
Solve: [latex]p+\Large\frac{1}{2}=\Large\frac{2}{3}[/latex]
Answer:
Solution:
|
[latex]p+\Large\frac{1}{2}=\Large\frac{2}{3}[/latex] |
| Subtract [latex]\Large\frac{1}{2}[/latex] from each side to undo the addition. |
[latex]p+\Large\frac{1}{2}\color{red}{-}\color{red}{\Large\frac{1}{2}} =\Large\frac{2}{3}\color{red}{-}\color{red}{\Large\frac{1}{2}}[/latex] |
| Simplify on each side of the equation. |
[latex]p + 0 =\Large\frac{2}{3}-\Large\frac{1}{2}[/latex] |
| Simplify the fraction by finding a common denominator. |
[latex]p =\Large\frac{2}{3}\cdot\color{red}{\Large\frac{2}{2}}-\Large\frac{1}{2}\cdot\color{red}{\Large\frac{3}{3}}[/latex] |
|
[latex]p =\Large\frac{4}{6}-\Large\frac{3}{6}[/latex] |
|
[latex]p =\Large\frac{1}{6}[/latex] |
| Check: |
[latex]p+\Large\frac{1}{2}=\Large\frac{2}{3}[/latex] |
|
| Substitute [latex]p=\Large\frac{1}{6}[/latex] . Rewrite as fractions with the LCD. |
[latex]\color{red}{\Large\frac{1}{6}} +\Large\frac{1}{2} \stackrel{?}{=}\Large\frac{2}{3}=\color{red}{\Large\frac{1}{6}}+\Large\frac{3}{6}\stackrel{?}{=}\Large\frac{4}{6}[/latex] |
|
|
[latex]\Large\frac{4}{6} =\Large\frac{4}{6}\quad\checkmark[/latex] |
|
We know we have found the solution to this equation since [latex]\Large\frac{4}{6} =\Large\frac{4}{6}[/latex].
Example
Solve: [latex]a-\Large\frac{1}{4}=-\Large\frac{2}{3}[/latex]
Answer:
Solution:
|
[latex]a-\Large\frac{1}{4}=-\Large\frac{2}{3}[/latex] |
| Add [latex]\Large\frac{5}{9}[/latex] from each side to undo the addition. |
[latex]a-\Large\frac{1}{4}\color{red}{+}\color{red}{\Large\frac{1}{4}} =-\Large\frac{2}{3}\color{red}{+}\color{red}{\Large\frac{1}{4}}[/latex] |
| Find a common denominator. |
[latex]a =-\Large\frac{2}{3}\cdot\color{red}{\Large\frac{4}{4}}+\Large\frac{1}{4}\color{red}{\Large\frac{3}{3}}[/latex] |
| Simplify on each side of the equation. |
[latex]a = -\Large\frac{8}{12}+\Large\frac{3}{12}[/latex] |
| Simplify the fraction. |
[latex]a = -\Large\frac{5}{12}[/latex] |
| Check: |
[latex]a-\Large\frac{1}{4}=-\Large\frac{2}{3}[/latex] |
|
| Substitute [latex]a=-\frac{5}{12}[/latex] . |
[latex]\color{red}{-\Large\frac{5}{12}} -\Large\frac{1}{4}\stackrel{?}{=}-\Large\frac{2}{3}[/latex] |
|
| Change to common denominator. |
[latex]\color{red}{-\Large\frac{5}{12}} -\Large\frac{3}{12}\stackrel{?}{=}-\Large\frac{8}{12}[/latex] |
|
| Subtract. |
[latex]-\Large\frac{8}{12} = -\Large\frac{8}{12}\quad\checkmark[/latex] |
|
Since [latex]a=-\Large\frac{5}{12}[/latex] makes the equation true, we know that [latex]a=-\Large\frac{5}{12}[/latex] is the solution to the equation.
