Algebra 2 Rational Expressions Worksheet Solutions (PDF)

Algebra 2 math worksheet solutions: Rational Expressions – Maths PDF

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This rational expressions answers for Algebra 2 covers rational expressions (Common Core A-APR.D.7, A-SSE.A.2), with a free PDF to print.

Every worksheet problem on rational expressions is solved here, one step at a time. Compare your method as well as your final answer, because the steps show where canceling terms instead of factors can go wrong. Final answers appear in bold so you can find them quickly when you only need a check.

Algebra 2 · Common Core State Standards: A-APR.D.7, A-SSE.A.2

Rational expressions answers: exercises 1 to 10

Exercise 1 – Excluded values

Set the denominator to 0: $x^{2} – 5x + 6 = (x – 2)(x – 3) = 0$.

$x \ne 2$ and $x \ne 3$

Exercise 2 – Simplify a monomial quotient

Factor: $\frac{2x(x + 3)}{4x} = \frac{x + 3}{2}$.

$\frac{x + 3}{2}$, $x \ne 0$

Exercise 3 – Simplify, trinomial denominator

$\frac{(x – 3)(x + 3)}{(x + 3)(x – 2)}$. Cancel $x + 3$.

$\frac{x – 3}{x – 2}$, $x \ne -3, 2$

Exercise 4 – Simplify, common factor x – 2

$\frac{(x – 2)(x – 3)}{(x – 2)(x + 2)}$.

$\frac{x – 3}{x + 2}$, $x \ne \pm 2$

Exercise 5 – Perfect square denominator

$\frac{(x – 4)(x + 4)}{(x + 4)^{2}}$.

$\frac{x – 4}{x + 4}$, $x \ne -4$

Exercise 6 – Hole in a graph

y = (x² – 4)/(x – 2) is the line y = x + 2 with a hole at x = 2.

$\frac{(x – 2)(x + 2)}{x – 2} = x + 2$ for $x \ne 2$.

The function is not defined at $x = 2$, so there is a hole at $(2, 4)$, where the line $y = x + 2$ would have been.

f(x) = x + 2 for x ≠ 2; hole at (2, 4)

Exercise 7 – Multiply

$x^{2} – 16 = (x – 4)(x + 4)$. Cancel $x + 3$ and $x – 4$.

$x + 4$

Exercise 8 – Multiply, factor first

$\frac{(x – 1)(x + 1)}{x + 3} \cdot \frac{x(x + 3)}{x + 1}$. Cancel $x + 1$ and $x + 3$.

$x(x – 1)$, $x \ne -3, -1$

Exercise 9 – Divide

Multiply by the reciprocal: $\frac{(x – 5)(x + 5)}{x + 1} \cdot \frac{(x – 1)(x + 1)}{x + 5}$.

$(x – 5)(x – 1)$

Exercise 10 – Add, unlike denominators

The LCD is $(x + 2)(x – 1)$.

$\frac{3(x – 1) + 5(x + 2)}{(x + 2)(x – 1)} = \frac{3x – 3 + 5x + 10}{(x + 2)(x – 1)}$.

$\frac{8x + 7}{(x + 2)(x – 1)}$

Exercises 11 and up

Exercise 11 – Subtract

LCD $(x – 3)(x + 3)$: $\frac{x(x + 3) – 2(x – 3)}{(x – 3)(x + 3)}$.

Numerator: $x^{2} + 3x – 2x + 6 = x^{2} + x + 6$.

$\frac{x^{2} + x + 6}{x^{2} – 9}$

Exercise 12 – Add with a factorable denominator

$x^{2} – 1 = (x – 1)(x + 1)$, so the LCD is $(x – 1)(x + 1)$.

$\frac{1 + (x + 1)}{(x – 1)(x + 1)}$.

$\frac{x + 2}{(x – 1)(x + 1)}$

Exercise 13 – Subtract and simplify

LCD $(x – 2)(x + 2)$: $\frac{x – (x + 2)}{(x – 2)(x + 2)}$.

$\frac{-2}{x^{2} – 4}$

Exercise 14 – Complex fraction

Multiply the numerator and denominator by $x^{2}$: $\frac{x^{2} – 4}{x^{2} – 2x} = \frac{(x – 2)(x + 2)}{x(x – 2)}$.

$\frac{x + 2}{x}$

Exercise 15 – Use structure

$16x^{4} – 81 = (4x^{2})^{2} – 9^{2} = (4x^{2} – 9)(4x^{2} + 9)$.

Cancel $4x^{2} + 9$.

$4x^{2} – 9 = (2x – 3)(2x + 3)$

Exercise 16 – Find the error

Only common factors cancel, not terms. The 3 in the numerator is added to $x$.

At $x = 6$: $\frac{6 + 3}{3} = 3$ but $6 + 1 = 7$. The expression cannot be simplified further.

Mei is wrong; (x + 3)/3 is already simplified

Exercise 17 – Painting together

Aisha paints 1/6 of the fence per hour and Diego 1/4 per hour: together 5/12.

Rates: $\frac{1}{6}$ and $\frac{1}{4}$ of the fence per hour.

Combined: $\frac{1}{6} + \frac{1}{4} = \frac{2}{12} + \frac{3}{12} = \frac{5}{12}$.

Time: $1 \div \frac{5}{12} = \frac{12}{5}$ hours.

2.4 hours (2 hours 24 minutes)

Exercise 18 – Average speed

Times: $\frac{60}{40} = 1.5$ hours and $\frac{60}{60} = 1$ hour.

Average speed $= \frac{120}{2.5} = 48$ mph, not the simple average 50.

48 mph

Exercise 19 – Lens equation

$\frac{1}{f} = \frac{1}{4} + \frac{1}{12} = \frac{3}{12} + \frac{1}{12} = \frac{4}{12} = \frac{1}{3}$.

$f = 3$ cm

Exercise 20 – Average cost

Average cost approaches 20.

$A(50) = \frac{500 + 1000}{50} = 30$; $A(100) = \frac{2500}{100} = 25$; $A(250) = \frac{5500}{250} = 22$.

$30, $25, $22

Exercise 21 – Equivalent expressions

$\frac{x^{2} – 9}{x – 3} = \frac{(x – 3)(x + 3)}{x – 3} = x + 3$ for $x \ne 3$.

At $x = 3$ the first expression has a denominator of 0 and is undefined, while $x + 3 = 6$.

Equal except at x = 3, where the first is undefined

Key facts behind this rational expressions answers

Key facts

Rational expression = polynomial ÷ polynomial.
Factor first, then cancel common factors only.
Excluded values make the original denominator 0.
Divide by multiplying by the reciprocal.
Add or subtract over the least common denominator.

How to use this rational expressions answers

Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.

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