Lesson 17: Solving Equations and Inequalities with Absolute Value (Exploration: Transforming f(x) = |x|)

New Concept

The absolute value of a number is the distance along the x-axis from the origin to the graph of the number.

What’s next

Next, you'll use this definition to solve equations and inequalities by breaking them into two distinct cases.

The absolute value of a number is its distance from the origin on the x-axis. It's always non-negative. If $x < 0$, then $|x| = -x$. If $x \geq 0$, then $|x| = x$.

For $|-8|$, since $-8 < 0$, the result is $-(-8) = 8$.
For $|12|$, since $12 > 0$, the result is simply $12$.
For $|0|$, the result is $0$, as it has no distance from itself.

Think of absolute value as a 'positivity machine.' It doesn't care if you have 50 dollars or a debt of 50 dollars, it only sees the amount: 50! The distance from zero is always positive. This is why the absolute value of any non-zero number, whether it's $-10$ or $10$, always comes out as a positive value.

To solve an absolute value equation like $|x - a| = k$, you must create and solve two separate cases derived from the original equation: $x - a = k$ or $x - a = -k$.

To solve $|x - 3| = 9$, you must solve both $x - 3 = 9$ (which gives $x=12$) and $x - 3 = -9$ (which gives $x=-6$).
To solve $|2y + 5| = 11$, you solve $2y + 5 = 11$ (getting $y=3$) and $2y + 5 = -11$ (getting $y=-8$).

Since absolute value makes everything positive, the expression inside could have started as either positive or negative. Imagine a mystery where the clue is 'the absolute value is 5.' The original number could have been $5$ or $-5$. We have to solve for both possibilities to crack the case and find all the hidden solutions!

Derived equations from an absolute value equation may result in extraneous solutions. These are solutions that do not satisfy the original absolute value equation. You must check all possible solutions by substituting them back into the original equation.

To solve $|2x + 6| = 4x$, we get two potential answers: $x=3$ and $x=-1$. Checking $x=3$ gives $|12|=12$, which is true. Checking $x=-1$ gives $|4|=-4$, which is false. Therefore, $x=-1$ is an extraneous solution.

Imagine following a recipe perfectly, but one ingredient was secretly expired! The final dish looks right, but it's not actually good. Extraneous solutions are like that—they pop up from your math steps but don't work when you plug them back into the original problem. Always check your answers to spot these impostors!