Example

Determine whether each of the following is a solution of [latex]x-\Large\frac{3}{10}=\Large\frac{1}{2}[/latex]

1. [latex]x=1[/latex]
2. [latex]x=\Large\frac{4}{5}[/latex]
3. [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.

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.

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.