Quick answer

Treat |expression| as non-negative. Split into cases when you solve or rewrite.

Formula

  • |x| ≥ 0
  • |ab| = |a|·|b|
  • |a/b| = |a|/|b| when b ≠ 0

Introduction

You will simplify |ax + b|, solve |ax + b| = c, and graph y = |ax + b| in the same chapter. Algebra ties the definition to systems you already know.

Case splits are not optional when the unknown sits inside bars. Each branch is a linear problem; combine results only after you check them.

The Absolute Value Calculator checks numeric pieces. Review absolute value equations and absolute value inequalities as you practice combined problems.

Where algebra uses | |

Piecewise definitions justify case splits. Product and quotient rules help when constants factor out of bars: |3y| = 3|y| when 3 is non-negative.

Word problems may ask for error bounds, tolerances, or symmetric intervals around a target value. Those stories often become |x − h| < k.

Do not distribute bars over addition without care: |a + b| is not always |a| + |b|. Look for special structure before you split.

Graphs, equations, and inequalities are three views of the same idea. Switch views when one method feels stuck.

Useful identities

  • |x| = max(−x, x)
  • |x − h| measures distance from h
  • If |x| &lt; d then −d &lt; x &lt; d
  • |ab| = |a|·|b|

Keep identities on a reference card. Use distance form when a problem mentions “within k units of h.”

When solving |ax + b| = c, isolate first, then split. When graphing, find the corner where ax + b = 0 before you plot.

Visual learners should connect identities to the V-shaped graph and to distance language on the number line.

Algebra workflow

  1. Simplify inside the bars when possible. Combine like terms before splitting cases unless the expression is already linear.
  2. Choose the right tool. Equal sign → two equations. Inequality → interval rewrite. Graph → table of values.
  3. Solve each case. Keep branch work separate to avoid mixing solutions.
  4. Verify and interpret. Check in the original statement and state units if given.

Combined skills

Solve |3x + 6| = 12. First 3x + 6 = 12 gives x = 2; second 3x + 6 = −12 gives x = −6. Both work in the original equation.

Rewrite |x − 4| < 2 as −2 < x − 4 < 2, so 2 < x < 6.

Graph y = |x + 1| by shifting the standard V one unit left; vertex at (−1, 0).