# 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.

$\left| {x + 2} \right| = – 3$

Since the absolute value can’t be negative, therefore it has $\color{red}\text{no solution}$.

Problem 2: Solve the absolute value equation below.

$\left| {2x \,- \,3} \right| = 7$

Solve the positive and negative cases.

Therefore, the solution set is $\{ – 2,5\}$.

Problem 3: Solve the absolute value equation below.

$– 7\left| {x\, – \,2} \right| = – 21$

Isolate the absolute value expression on the left side by dividing both sides of the equation by $-7$

\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.

$4\, – \,\left| {3x\,+ \,6} \right| = 1$

Isolate the absolute value expression by subtracting $4$ from both sides. Then, divide $-1$ 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.

$\left| {x \,- \,{1 \over 2}} \right| + \left| {x \, – \,{1 \over 2}} \right| = 7$

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 $2$ 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.

$\left| {3{x^2}\, – \,2x} \right| = 5$

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.

${\Large\left| {{{x \,- \,2} \over {x\, + \,1}}} \right|} ={ \Large{1 \over 2}}$

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.

$\left| {x\, – \,8} \right| = 2x\, -\, 1$

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.

$\left| {2x + 5} \right| = x + 7$

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.

$\left| {2x + 20} \right| = \left| {3x – 5} \right|$

Set up the equations then solve.

Therefore, the solution set is $\{ -3,25\}$.

Problem 11: Solve the absolute value equation below.

$\left| { – 9x + 10} \right| = \left| {11x} \right|$

Therefore, the solution set is $\{ -5,1/2\}$.