Absolute Value Equations
Solve absolute value equations by setting up two cases (the expression inside equals the positive and negative target values). Identify extraneous solutions and apply to distance-based problems.
A lesson of Algebra 1, in Linear Inequalities & Absolute Value, about 30 minutes.
By the end of this lesson, you will be able to solve absolute value equations by splitting them into two cases, recognize the equations that have one solution or none, and check every answer in the original equation.
Absolute Value Is a Distance, So There Are Two Answers
$$|x| = 5$$
Read that as a question about distance: which numbers sit exactly $5$ steps from zero?
$$-5 \quad \text{and} \quad 5$$
Two of them, one on each side. That is the whole idea of this lesson, and every technique below is a consequence of it.
Distance has no direction, so $|-5|$ and $|5|$ are both $5$. And distance is never negative, which is the reason some of these equations turn out to have no solutions at all.
The Two-Case Method
For any expression $A$ and any positive number $c$:
$$|A| = c \quad \text{means} \quad A = c \quad \text{or} \quad A = -c$$
Whatever is inside the bars is $c$ steps from zero, so it is either $c$ or $-c$. Solve both, and both answers count.
Two situations where it is not two:
| The equation | Why | How many solutions |
|---|---|---|
| $\lvert A \rvert = 0$ | Only zero is zero steps from zero | One, from $A = 0$ |
| $\lvert A \rvert = -3$ | A distance is never negative | None |
The second one is worth spotting immediately. If the right hand side is negative once the bars are alone, you are finished, and no case work is needed.
Example 1: The Standard Two Cases
Here is the whole method on one equation, worked all the way through. Solve $|2x - 3| = 7$. Whatever $2x - 3$ is, it is $7$ steps from zero. So it is $7$, or it is $-7$.
- Write the first case and solve it x = 5
- Write the second case and solve it x = -2
- Check the first solution. Work out the left hand side of the original at x = 5 7
- At x = -2, what does 2x - 3 come to, BEFORE taking the absolute value? -7
Worked Example 1: Your Turn
Solve $|3x + 1| = 10$.
Example 2: Get the Bars Alone First
You cannot split into cases while anything is still attached to the outside of the absolute value. Solve $|3x + 6| + 4 = 10$.
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