Absolute Value Equations Practice Problems with Answers
There are eleven (11) practice problems in this collection regarding absolute value equations. Whether you’re a beginner or seeking a challenge, there’s something here for you. As you tackle each problem, remember that in every attempt, right or wrong, fuels your mathematical growth. Good luck!
Problem 1: Solve the absolute value equation below.
\(\left| {x + 2} \right| = – 3\)
Answer
Since the absolute value can’t be negative, therefore it has \(\color{red}\text{no solution}\).
Problem 2: Solve the absolute value equation below.
\(\left| {2x \,- \,3} \right| = 7\)
Answer
Solve the positive and negative cases.
Therefore, the solution set is \(\{ – 2,5\}\).
Problem 3: Solve the absolute value equation below.
\( – 7\left| {x\, – \,2} \right| = – 21\)
Answer
Isolate the absolute value expression on the left side by dividing both sides of the equation by [latex]-7[/latex]
\begin{align*} – 7\left| {x – 2} \right| = – 21 \\ \\ {{ – 7\left| {x – 2} \right|} \over { – 7}} = {{ – 21} \over { – 7}} \\ \\ \left| {x – 2} \right| = 3 \end{align*}Now, break the absolute value equation into two cases then solve.
Therefore, the solution set is \(\{ – 1,5\}\).
Problem 4: Solve the absolute value equation below.
\(4\, – \,\left| {3x\,+ \,6} \right| = 1\)
Answer
Isolate the absolute value expression by subtracting [latex]4[/latex] from both sides. Then, divide [latex]-1[/latex] from both sides.
\begin{align*} 4 – \left| {3x + 6} \right| &= 1 \\ \\ 4 – 4 – \left| {3x + 6} \right| &= 1 – 4 \\ \\ – \left| {3x + 6} \right| &= – 3 \\ \\ {{ – \left| {3x + 6} \right|} \over { – 1}} &= {{ – 3} \over { – 3}} \\ \\ \left| {3x + 6} \right| &= 3 \end{align*}Break the absolute value equation into two equations with positive and negative cases then solve.
Therefore, the solution set is \(\{ – 3,-1\}\).
Problem 5: Solve the absolute value equation below.
\(\left| {x \,- \,{1 \over 2}} \right| + \left| {x \, – \,{1 \over 2}} \right| = 7\)
Answer
The absolute value expressions on the left side are similar terms therefore we can combine them.
\begin{align*} \left| {x \,-\, {1 \over 2}} \right| + \left| {x \,- \,{1 \over 2}} \right| &= 7 \\ \\ 2\left| {x\, – \,{1 \over 2}} \right| &= 7 \end{align*}Finally, divide [latex]2[/latex] from sides of the equation to isolate the absolute value expression.
\begin{align*} & {{2\left| {x\, -\, {1 \over 2}} \right|} \over 2} = {7 \over 2} \\ \\ & \left| {x \,- \,{1 \over 2}} \right| = {7 \over 2} \end{align*}Now, we solve each case.
Therefore, the solution set is \(\{ – 3,4\}\).
Problem 6: Solve the absolute value equation below.
\(\left| {3{x^2}\, – \,2x} \right| = 5\)
Answer
Split the absolute value equation into two cases then solve.
Therefore, the solution set is \(\{ – 1,{\large{5 \over 3}}\}\).
Problem 7: Solve the absolute value equation below.
\({\Large\left| {{{x \,- \,2} \over {x\, + \,1}}} \right|} ={ \Large{1 \over 2}}\)
Answer
Separate the absolute value equation into positive and negative cases then solve.
Therefore, the solution set is \(\{ 1,5\}\).
Problem 8: Solve the absolute value equation below.
\(\left| {x\, – \,8} \right| = 2x\, -\, 1\)
Answer
Consider the positive and negative cases then solve each equation.
Validate the answer \(x=-7\) if it is indeed a solution by plugging it back into the original absolute value equation.
We arrived at a contradiction since \(15=-15\) is false. That means \(x=-7\) is not a solution.
Let’s also validate \(x=3\).
It yields a true statement which means \(x=3\) is a solution.
Therefore, the solution set is \(\{3\}\).
Problem 9: Solve the absolute value equation below.
\(\left| {2x + 5} \right| = x + 7\)
Answer
Break the absolute value equation into two cases of linear equations the solve.
Substitute \(x=2\)back into the original absolute value equation to check if it’s true or not. If true, then it is a solution. If false, then it is not.
It checks! Therefore, \(x=2\) is a solution.
We also validate \(x=-4\) using the same procedure as above.
Since it gives a true statement that means \(x=-4\) is also a solution.
Therefore, the solution set is \(\{ -4,2\}\).
Problem 10: Solve the absolute value equation below.
\(\left| {2x + 20} \right| = \left| {3x – 5} \right|\)
Answer
Set up the equations then solve.
Therefore, the solution set is \(\{ -3,25\}\).
Problem 11: Solve the absolute value equation below.
\(\left| { – 9x + 10} \right| = \left| {11x} \right|\)
Answer
Set up the absolute value equation into cases then solve.
Therefore, the solution set is \(\{ -5,1/2\}\).
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