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Rational Expressions Worksheet Answer Key – Algebra 2

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Last updated: October 22, 2026 · Written by the Math Made Easy math teachers

Algebra 2 Math Worksheet Answer Keys – Mathbot, the Math Made Easy robot coach, with free printable worksheet answer keys for Algebra 2

Looking for a rational expressions worksheet answer key? Then this answer key gives every step of the Algebra 2 worksheet on Rational Expressions and Equations Step by Step.

Each solution factors first, names the common denominator, and checks the final answers against the excluded values. Compare each step with the method in the lesson.

Standards: A-APR.D.6, A-APR.D.7, A-REI.A.2 · Level: Grade 11 Algebra 2

Problem 2 Excluded values

a) x2 − 9 = (x − 3)(x + 3), so x ≠ 3 and x ≠ −3.

b) 2x − 7 = 0 at x = 72.

c) x2 + x − 12 = (x + 4)(x − 3), so x ≠ −4 and x ≠ 3.

(a) ±3; (b) 72; (c) −4 and 3

Problem 3 Simplifying by factoring

Numerator: (x − 5)(x + 5). Denominator: (x − 5)(x + 2).

Cancel x − 5: x + 5x + 2, with x ≠ 5 and x ≠ −2.

Problem 4 Multiplying rational expressions

The factor x + 5 cancels, and 3x2/(9x) = x/3.

x3, with x ≠ −5 and x ≠ 0.

Problem 5 Excluded values on a number line

x2 − x − 20 = (x − 5)(x + 4) = 0 at x = 5 and x = −4.

-8-7-6-5-4-3-2-1012345678−45
Excluded values −4 and 5.

Domain: all real numbers except −4 and 5

Problem 6 Dividing rational expressions

Multiply by the reciprocal: (x − 2)(x + 2)x + 3 · x(x + 3)x + 2.

Cancel x + 2 and x + 3 to get x(x − 2), with x ≠ −3, −2, 0.

x(x − 2)

Problem 7 Like denominators

a) 3x + 1 + 2x − 9x − 4 = 5x − 8x − 4.

b) Distribute the minus: 7x + 2 − x + 10x + 6 = 6x + 12x + 6 = 6(x + 2)x + 6.

Problem 8 Unlike denominators

The LCD is (x + 5)(x − 3).

7(x − 3) + 2(x + 5)(x + 5)(x − 3) = 7x − 21 + 2x + 10(x + 5)(x − 3).

9x − 11(x + 5)(x − 3)

Problem 9 Factoring a denominator first

x2 − 9 = (x − 3)(x + 3), so the LCD is (x − 3)(x + 3).

5 − 2(x + 3)(x − 3)(x + 3) = 5 − 2x − 6(x − 3)(x + 3).

−2x − 1(x − 3)(x + 3)

Problem 10 Adding with a quadratic denominator

x2 − 5x + 6 = (x − 2)(x − 3), so the second fraction needs the factor x − 3.

x + 4(x − 3)(x − 2)(x − 3) = x + 4x − 12(x − 2)(x − 3).

5x − 12(x − 2)(x − 3)

Problem 11 A first rational equation

Multiply every term by x: 8 − 3x = 2, so 3x = 6 and x = 2.

Check: 8/2 − 3 = 1 and 2/2 = 1 ✓.

x = 2

Problem 12 No solution

Multiply by x − 5: x + 2(x − 5) = 5, so 3x = 15 and x = 5.

But x = 5 makes the denominator 0, so it is extraneous.

No solution

Problem 13 Solving with a graph

The graphs cross at (3, 2) and (−2, −3).

Algebra: multiply by x: 6 = x2 − x, so x2 − x − 6 = 0 and (x − 3)(x + 2) = 0.

-6-5-4-3-2-11234567-6-5-4-3-2-11234567xy(3, 2)(-2, -3)
Both solutions are valid.

x = 3 and x = −2

Problem 14 Two denominators and a quadratic

Multiply by (x − 1)(x + 4): 3(x + 4) + 2(x − 1) = (x − 1)(x + 4).

5x + 10 = x2 + 3x − 4, so x2 − 2x − 14 = 0.

x = 2 ± √(4 + 56)2 = 2 ± 2√152 = 1 ± √15.

Neither value equals 1 or −4, so both are valid.

x = 1 + √15 or x = 1 − √15

Problem 15 Rejecting an extraneous solution

The LCD is (x − 2)(x + 2). Multiply: 3(x + 2) − 12 = x2 − 4, so 3x − 6 = x2 − 4 and x2 − 3x + 2 = 0.

(x − 1)(x − 2) = 0, so x = 1 or x = 2.

x = 2 makes a denominator 0, so reject it. Check x = 1: −3 + 4 = 1 ✓.

x = 1

Problem 16 Painting a fence together

Ana: 16 per hour. Ben: 19 per hour. Together: 16 + 19 = 318 + 218 = 518 per hour.

Time = 1 ÷ 518 = 185 = 3.6 hours.

Ana: 1/6 per hourBen: 1/9 per hourTogether: 5/18 per houreach bar = the whole job
Together, 3.6 hours fill the whole job.

3.6 hours (3 hours 36 minutes)

Problem 17 Two hoses filling a pond

Let t be the hours for the faster hose; the slower takes t + 5. Rates: 1t + 1t + 5 = 16.

Multiply by 6t(t + 5): 6(t + 5) + 6t = t(t + 5), so 12t + 30 = t2 + 5t.

t2 − 7t − 30 = 0, which factors as (t − 10)(t + 3) = 0. The time must be positive, so t = 10.

Check: 1/10 + 1/15 = 5/30 = 1/6 ✓.

Faster hose 10 hours; slower hose 15 hours

Problem 18 Paddling against a current

Let s be the still-water speed. Upstream speed s − 3, downstream s + 3. Total time:

12s − 3 + 12s + 3 = 3.

Multiply by (s − 3)(s + 3): 12(s + 3) + 12(s − 3) = 3(s2 − 9), so 24s = 3s2 − 27 and s2 − 8s − 9 = 0.

(s − 9)(s + 1) = 0. A speed must be positive, so s = 9.

Check: 12/6 + 12/12 = 2 + 1 = 3 ✓.

9 mph

Problem 19 Resistors in parallel

1x + 1x + 6 = 14. Multiply by 4x(x + 6): 4(x + 6) + 4x = x(x + 6).

8x + 24 = x2 + 6x, so x2 − 2x − 24 = 0 and (x − 6)(x + 4) = 0. A resistance is positive, so x = 6.

6 Ω12 Ω6Vtwo resistors in parallel
Resistances 6 and 12 ohms.

Check: 1/6 + 1/12 = 3/12 = 1/4 ✓.

6 ohms and 12 ohms

Problem 20 A multi-step quotient

Rewrite as a product: (x − 3)(x + 3)x(x + 2) · (x − 2)(x + 2)x + 3.

Cancel x + 3 and x + 2: (x − 3)(x − 2)x.

Excluded: x = 0 and x = −2 (original denominator), x = −3 (divisor is 0), and x = ±2 (from x2 − 4).

(x − 3)(x − 2)x, with x ≠ 0, −2, −3, 2