Lesson 7 of 12 · Algebra Foundations
Rational Expressions & Absolute Value
Two topics, one shared habit: factor before you do anything else. Rational expressions simplify the same way numeric fractions do, once you factor numerator and denominator — and absolute value is nothing more than a distance, which turns every equation and inequality into a simple two-case (or two-boundary) statement.
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Learning objectives
- Simplify rational expressions by factoring numerator and denominator, tracking which values must be excluded from the domain.
- Add, subtract, multiply, and divide rational expressions using a common denominator.
- Solve absolute value equations by splitting into two cases.
- Solve and interpret absolute value inequalities as distance statements — "less than" means between two boundaries, "greater than" means outside them.
The core idea
A rational expression is a fraction whose numerator and denominator are polynomials. Every technique you already know for numeric fractions applies — but you must factor first, because "canceling" only makes sense between actual common factors, not between terms that merely look similar. Factoring also reveals the expression's domain: any x-value that makes the original denominator zero is excluded, even if that factor later cancels.
Absolute value has a single geometric meaning: |x| is the distance from x to 0, and more generally |x − a| is the distance between x and a. Once you read it that way, absolute value equations and inequalities stop being special cases to memorize and become ordinary distance statements.
Simplifying a rational expression
Factor the numerator and denominator completely, cancel common factors, and separately record every x-value that made the original denominator zero — those stay excluded from the domain regardless of what cancels.
Absolute value equation
|ax + b| = c (c ≥ 0) ⟺ ax + b = c or ax + b = −c
Absolute value inequality — "less than" (between)
|x − a| < b ⟺ a − b < x < a + b
"Distance from a is less than b" means x is trapped between a−b and a+b.
Absolute value inequality — "greater than" (outside)
|x − a| > b ⟺ x < a − b or x > a + b
"Distance from a is more than b" means x is outside the interval, on either side.
Worked example 1 — Simplifying with a hidden excluded value
Problem
Simplify (x² − 4)/(x² − x − 6) and state every excluded value.
Key insight
Factor both top and bottom before canceling — and note which factors were in the original denominator.
Solution
(x−2)(x+2) / [(x−3)(x+2)] = (x−2)/(x−3). The original denominator was zero at x = 3 and x = −2, so both are excluded — even though (x+2) canceled.
Takeaway
The simplified expression and the original expression are not quite identical functions — they differ at the canceled value, which is exactly why that value must still be excluded.
Worked example 2 — Combining fractions
Problem
Write 1/(x−1) + 2/(x+1) as a single fraction.
Key insight
Use the common denominator (x−1)(x+1), exactly like combining numeric fractions.
Solution
[(x+1) + 2(x−1)] / [(x−1)(x+1)] = (3x−1)/(x²−1).
Takeaway
Expand the numerator carefully — distributing the 2 across (x−1) is the step most likely to have a sign slip.
Worked example 3 — Absolute value equation
Problem
Solve |2x − 3| = 7.
Key insight
The expression inside could be +7 or −7 — both give valid equations to solve.
Solution
2x−3=7 → x=5, or 2x−3=−7 → x=−2. x = 5 or −2.
Takeaway
Every absolute value equation |expr|=k (k>0) has exactly two cases — never forget the negative one.
Worked example 4 — Absolute value inequality as distance
Problem
Solve |x − 4| < 3 and describe the solution as a distance statement.
Key insight
"|x − 4| < 3" literally says "the distance from x to 4 is less than 3."
Solution
That means x is within 3 of 4 in either direction: 4−3 < x < 4+3, i.e. 1 < x < 7.
Takeaway
You never need to split |x−4|<3 into cases and solve each — reading it as a distance gives the interval directly.
Strategy notes
Factor before you cancel — always
It's tempting to cancel individual terms that "look the same" between numerator and denominator. Only fully factored common factors can be canceled — never terms within a sum.
Translate absolute value into a distance sentence
Before doing any algebra, read |x − a| as "distance from x to a." This single habit makes the difference between "less than = between" and "greater than = outside" automatic instead of memorized.
Common mistakes
Forgetting an excluded value that canceled away
If a factor in the denominator cancels, the value that made it zero is still excluded from the domain — the simplified expression is only equal to the original everywhere except that point.
Only solving one case of an absolute value equation
|expr| = k always has two cases (for k > 0): expr = k and expr = −k. Solving only the "obvious" positive case is one of the most common point losses on this topic.
Flipping "less than" and "greater than" for inequalities
|x−a| < b is a bounded interval ("between"); |x−a| > b is two unbounded rays ("outside"). Mixing these up is easy under time pressure — anchor to the distance interpretation to avoid it.
Practice
20 questions across four difficulty tiers. Simplification questions want your simplified expression (any equivalent form). Equation questions want comma-separated values. Inequality questions ask for the two boundary values of the interval, not the inequality itself. Up to three tries per question before the solution is shown.