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

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 1 Simplifying monomial fractions
Divide the numbers by 6 and cancel powers: 12/18 = 2/3, x3/x = x2, y/y2 = 1/y.
2x23y
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.
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.
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.
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.
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
