L-ALG-4: Rational Expressions — Solutions

Full worked solutions · Sections 5–9

Solutions are provided for all problems from Sections 5–9. Generator problems (Section 6, P1 and P2) produce their own inline solutions — use the "Generate new problem" button for fresh problems with solutions.

← Back to Lesson

Section 5 Solutions — Guided Practice

▾

A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Problem 1 — Simplify and Find the Domain (Variant Bank)

Variant 0: \( \dfrac{x^2-4}{x^2+x-2} \)

Factor: \( x^2-4 = (x-2)(x+2) \), \( x^2+x-2 = (x+2)(x-1) \).

Domain: \( x \neq 1, x \neq -2 \) (from original denominator).

Cancel \( (x+2) \): \( \dfrac{x-2}{x-1}, \quad x \neq 1, \; x \neq -2 \)

Variant 1: \( \dfrac{x^2-9}{x^2-x-6} \)

Factor: \( x^2-9 = (x-3)(x+3) \), \( x^2-x-6 = (x-3)(x+2) \).

Domain: \( x \neq 3, x \neq -2 \).

Cancel \( (x-3) \): \( \dfrac{x+3}{x+2}, \quad x \neq 3, \; x \neq -2 \)

Variant 2: \( \dfrac{x^2-x-12}{x^2-6x+8} \)

Factor: \( x^2-x-12 = (x-4)(x+3) \), \( x^2-6x+8 = (x-4)(x-2) \).

Domain: \( x \neq 4, x \neq 2 \).

Cancel \( (x-4) \): \( \dfrac{x+3}{x-2}, \quad x \neq 4, \; x \neq 2 \)

Variant 3: \( \dfrac{x^2+2x-3}{x^2-1} \)

Factor: \( x^2+2x-3 = (x+3)(x-1) \), \( x^2-1 = (x-1)(x+1) \).

Domain: \( x \neq 1, x \neq -1 \).

Cancel \( (x-1) \): \( \dfrac{x+3}{x+1}, \quad x \neq 1, \; x \neq -1 \)

Variant 4: \( \dfrac{x^2-5x+6}{x^2-4x+3} \)

Factor: \( x^2-5x+6 = (x-2)(x-3) \), \( x^2-4x+3 = (x-1)(x-3) \).

Domain: \( x \neq 3, x \neq 1 \).

Cancel \( (x-3) \): \( \dfrac{x-2}{x-1}, \quad x \neq 3, \; x \neq 1 \)

Problem 2 — Building the LCD (Non-regenerable)

Add \( \dfrac{5}{x} + \dfrac{2}{x+3} + \dfrac{1}{x^2+3x} \).

Factor \( x^2+3x = x(x+3) \). LCD \( = x(x+3) \).

\[ \frac{5(x+3)}{x(x+3)} + \frac{2x}{x(x+3)} + \frac{1}{x(x+3)} = \frac{5x+15+2x+1}{x(x+3)} = \frac{7x+16}{x(x+3)} \]

Check if \( 7x+16 \) shares a factor with \( x(x+3) \): at \( x=0 \), \( 7(0)+16=16 \neq 0 \); at \( x=-3 \), \( 7(-3)+16=-5 \neq 0 \). Not factorable further.

\[ \frac{7x+16}{x(x+3)}, \quad x \neq 0, \; x \neq -3 \]

Problem 3 — Subtraction Sign Distribution (Variant Bank)

Variant 0: \( \dfrac{4}{x+1} - \dfrac{2}{x-3} \)

LCD \( = (x+1)(x-3) \). Numerator: \( 4(x-3) - 2(x+1) = 4x-12-2x-2 = 2x-14 = 2(x-7) \).

\[ \frac{2(x-7)}{(x+1)(x-3)}, \quad x \neq -1, \; x \neq 3 \]

Variant 1: \( \dfrac{5}{x-2} - \dfrac{3}{x+4} \)

LCD \( = (x-2)(x+4) \). Numerator: \( 5(x+4) - 3(x-2) = 5x+20-3x+6 = 2x+26 = 2(x+13) \).

\[ \frac{2(x+13)}{(x-2)(x+4)}, \quad x \neq 2, \; x \neq -4 \]

Variant 2: \( \dfrac{3}{x+5} - \dfrac{1}{x-1} \)

LCD \( = (x+5)(x-1) \). Numerator: \( 3(x-1) - 1(x+5) = 3x-3-x-5 = 2x-8 = 2(x-4) \).

\[ \frac{2(x-4)}{(x+5)(x-1)}, \quad x \neq -5, \; x \neq 1 \]

Variant 3: \( \dfrac{6}{x-4} - \dfrac{2}{x+3} \)

LCD \( = (x-4)(x+3) \). Numerator: \( 6(x+3) - 2(x-4) = 6x+18-2x+8 = 4x+26 = 2(2x+13) \).

\[ \frac{2(2x+13)}{(x-4)(x+3)}, \quad x \neq 4, \; x \neq -3 \]

Variant 4: \( \dfrac{7}{x+2} - \dfrac{4}{x-5} \)

LCD \( = (x+2)(x-5) \). Numerator: \( 7(x-5) - 4(x+2) = 7x-35-4x-8 = 3x-43 \).

\[ \frac{3x-43}{(x+2)(x-5)}, \quad x \neq -2, \; x \neq 5 \]

Problem 4 — Complex Fraction (Non-regenerable)

Simplify \( \dfrac{\dfrac{x+1}{x-3}}{\dfrac{x^2-1}{x^2-9}} \).

Rewrite as division: \( \dfrac{x+1}{x-3} \div \dfrac{x^2-1}{x^2-9} = \dfrac{x+1}{x-3} \cdot \dfrac{x^2-9}{x^2-1} \)

Factor all: \( x^2-1 = (x-1)(x+1) \), \( x^2-9 = (x-3)(x+3) \).

\[ \frac{(x+1)}{(x-3)} \cdot \frac{(x-3)(x+3)}{(x-1)(x+1)} = \frac{\cancel{(x+1)}\cancel{(x-3)}(x+3)}{\cancel{(x-3)}(x-1)\cancel{(x+1)}} = \frac{x+3}{x-1} \]

Domain: \( x \neq 3, \; x \neq -3, \; x \neq 1, \; x \neq -1 \)

Section 6 Solutions — Independent Practice

▾

A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Problems 1 and 2 (generators) produce inline solutions. See below for the variant bank and non-regenerable problems.

Problem 3 — Multiply or Divide (Variant Bank)

Variant 0: \( \dfrac{x^2-1}{x+2} \cdot \dfrac{x+2}{x+1} \)

\( = \dfrac{(x-1)(x+1)}{x+2} \cdot \dfrac{x+2}{x+1} = \dfrac{(x-1)\cancel{(x+1)}\cancel{(x+2)}}{\cancel{(x+2)}\cancel{(x+1)}} = x-1 \)

Domain: \( x \neq -2, \; x \neq -1 \)

Variant 1: \( \dfrac{x^2-4}{x-3} \cdot \dfrac{x-3}{x+2} \)

\( = \dfrac{(x-2)(x+2)}{x-3} \cdot \dfrac{x-3}{x+2} = x-2 \)

Domain: \( x \neq 3, \; x \neq -2 \)

Variant 2: \( \dfrac{x^2-9}{x^2-x-6} \div \dfrac{x-3}{x+2} \)

Flip and multiply: \( \dfrac{(x-3)(x+3)}{(x-3)(x+2)} \cdot \dfrac{x+2}{x-3} = \dfrac{x+3}{x-3} \)

Domain: \( x \neq 3, \; x \neq -2 \)

Variant 3: \( \dfrac{2x^2+6x}{x^2-4} \cdot \dfrac{x-2}{x+3} \)

\( = \dfrac{2x(x+3)}{(x-2)(x+2)} \cdot \dfrac{x-2}{x+3} = \dfrac{2x}{x+2} \)

Domain: \( x \neq 2, \; x \neq -2, \; x \neq -3 \)

Variant 4: \( \dfrac{x^2+5x+6}{x^2-x-6} \cdot \dfrac{x-3}{x+2} \)

\( = \dfrac{(x+2)(x+3)}{(x-3)(x+2)} \cdot \dfrac{x-3}{x+2} = \dfrac{x+3}{x+2} \)

Domain: \( x \neq 3, \; x \neq -2 \)

Problem 4 — Add with Factoring Required (Variant Bank)

Variant 0: \( \dfrac{1}{x+2} + \dfrac{3}{x^2-4} \)

\( x^2-4 = (x-2)(x+2) \). LCD \( = (x-2)(x+2) \).

\( \dfrac{x-2+3}{(x-2)(x+2)} = \dfrac{x+1}{(x-2)(x+2)}, \quad x \neq 2, \; x \neq -2 \)

Variant 1: \( \dfrac{2}{x-3} + \dfrac{5}{x^2-9} \)

\( x^2-9 = (x-3)(x+3) \). LCD \( = (x-3)(x+3) \).

\( \dfrac{2(x+3)+5}{(x-3)(x+3)} = \dfrac{2x+11}{(x-3)(x+3)}, \quad x \neq 3, \; x \neq -3 \)

Variant 2: \( \dfrac{3}{x+1} + \dfrac{2}{x^2-1} \)

\( x^2-1 = (x-1)(x+1) \). LCD \( = (x-1)(x+1) \).

\( \dfrac{3(x-1)+2}{(x-1)(x+1)} = \dfrac{3x-1}{(x-1)(x+1)}, \quad x \neq 1, \; x \neq -1 \)

Variant 3: \( \dfrac{4}{x-2} + \dfrac{1}{x^2-x-2} \)

\( x^2-x-2 = (x-2)(x+1) \). LCD \( = (x-2)(x+1) \).

\( \dfrac{4(x+1)+1}{(x-2)(x+1)} = \dfrac{4x+5}{(x-2)(x+1)}, \quad x \neq 2, \; x \neq -1 \)

Variant 4: \( \dfrac{5}{x+4} + \dfrac{2}{x^2+x-12} \)

\( x^2+x-12 = (x+4)(x-3) \). LCD \( = (x+4)(x-3) \).

\( \dfrac{5(x-3)+2}{(x+4)(x-3)} = \dfrac{5x-13}{(x+4)(x-3)}, \quad x \neq -4, \; x \neq 3 \)

Problem 5 — Multi-Step Simplification (Non-regenerable)

\( \dfrac{x}{x^2-1} - \dfrac{1}{x^2+x} \)

Factor: \( x^2-1 = (x-1)(x+1) \), \( x^2+x = x(x+1) \). LCD \( = x(x-1)(x+1) \).

\[ \frac{x \cdot x}{x(x-1)(x+1)} - \frac{1 \cdot (x-1)}{x(x-1)(x+1)} = \frac{x^2 - (x-1)}{x(x-1)(x+1)} = \frac{x^2-x+1}{x(x-1)(x+1)} \]

Check factorability of \( x^2-x+1 \): discriminant \( = 1-4 = -3 < 0 \). Fully simplified.

\[ \frac{x^2-x+1}{x(x-1)(x+1)}, \quad x \neq 0, \; x \neq 1, \; x \neq -1 \]

Section 7 Solutions — Mastery Check

▾

A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Question 1 — Feynman Test (Model Answer)

In \( \dfrac{5x}{x} \), the denominator \( x \) is a factor of the entire numerator: \( 5x = x \cdot 5 \). We can rewrite as \( \dfrac{x \cdot 5}{x \cdot 1} \) and cancel the shared factor \( x \), leaving \( 5 \).

In \( \dfrac{x+5}{x} \), the \( x \) in the denominator is not a factor of the entire numerator \( x + 5 \) — it's only a factor of the first term. Since we cannot write \( x+5 = x \cdot (\text{polynomial}) \), cancellation is invalid. The fundamental rule: cancel factors (multiplicative), never terms (additive).

Question 2 — Holes vs. Asymptotes

\( f(x) = \dfrac{x^2-4}{x^2+x-2} \)

Factor: \( x^2-4 = (x-2)(x+2) \), \( x^2+x-2 = (x+2)(x-1) \).

Original function undefined at \( x = -2 \) and \( x = 1 \).

Simplified: \( \dfrac{x-2}{x-1} \), \( x \neq -2, \; x \neq 1 \).

At \( x = -2 \): simplified gives \( \dfrac{-4}{-3} = \dfrac{4}{3} \) — finite → hole at \( \left(-2, \dfrac{4}{3}\right) \).

At \( x = 1 \): simplified gives \( \dfrac{-1}{0} \) — still undefined → vertical asymptote at \( x = 1 \).

Question 3 — Error Analysis

The error is in expanding \( -1 \cdot (x+2) \). The student wrote \( -x + 2 \) (positive 2) instead of \( -x - 2 \).

Corrected: \( 3(x-1) - 1(x+2) = 3x-3-x-2 = 2x-5 \).

\[ \frac{3}{x+2} - \frac{1}{x-1} = \frac{2x-5}{(x+2)(x-1)} \]

Section 8 Solutions — Boss Fight

▾

A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Path A — Dissecting a Rational Function

\( f(x) = \dfrac{x^2+x-6}{x^2-x-2} = \dfrac{(x+3)(x-2)}{(x-2)(x+1)} \)

Domain: \( x \neq 2, \; x \neq -1 \) (from original denominator).

Cancel \( (x-2) \): simplified form \( \dfrac{x+3}{x+1} \).

At \( x = 2 \): \( \dfrac{2+3}{2+1} = \dfrac{5}{3} \) — finite → hole at \( \left(2, \dfrac{5}{3}\right) \).

At \( x = -1 \): \( \dfrac{-1+3}{-1+1} = \dfrac{2}{0} \) — undefined → vertical asymptote at \( x = -1 \).

Path B — Building a Derivative via the Difference Quotient

\( f(x) = \dfrac{3}{x+1} \), compute \( f'(x) \).

\( f(x+h) = \dfrac{3}{x+h+1} \)

\[ f(x+h) - f(x) = \frac{3}{x+h+1} - \frac{3}{x+1} = \frac{3(x+1) - 3(x+h+1)}{(x+h+1)(x+1)} \]

Numerator: \( 3x+3-3x-3h-3 = -3h \).

\[ \frac{f(x+h)-f(x)}{h} = \frac{-3h}{(x+h+1)(x+1) \cdot h} = \frac{-3}{(x+h+1)(x+1)} \]

\[ f'(x) = \lim_{h \to 0} \frac{-3}{(x+h+1)(x+1)} = \frac{-3}{(x+1)^2} \]

Section 9 Solutions — Challenge Problems

▾

A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Challenge 1 — Partial Fractions

Find \( A, B \) such that \( \dfrac{7x-1}{x^2-1} = \dfrac{A}{x-1} + \dfrac{B}{x+1} \).

Multiply both sides by \( (x-1)(x+1) \): \( 7x-1 = A(x+1) + B(x-1) \).

Set \( x=1 \): \( 6 = 2A \Rightarrow A = 3 \). Set \( x=-1 \): \( -8 = -2B \Rightarrow B = 4 \).

\[ \frac{7x-1}{x^2-1} = \frac{3}{x-1} + \frac{4}{x+1} \]

Verify: \( \dfrac{3(x+1)+4(x-1)}{x^2-1} = \dfrac{7x-1}{x^2-1} \) ✓

Challenge 2 — Three-Factor LCD

\( \dfrac{2x}{x^2-1} - \dfrac{3}{x^2+x} \)

Factor: \( x^2-1 = (x-1)(x+1) \), \( x^2+x = x(x+1) \). LCD \( = x(x-1)(x+1) \).

\[ \frac{2x^2}{x(x-1)(x+1)} - \frac{3(x-1)}{x(x-1)(x+1)} = \frac{2x^2-3x+3}{x(x-1)(x+1)} \]

Discriminant of \( 2x^2-3x+3 \): \( 9-24 = -15 < 0 \). Fully simplified.

\[ \frac{2x^2-3x+3}{x(x-1)(x+1)}, \quad x \neq 0, \; x \neq 1, \; x \neq -1 \]

Challenge 3 — Why Term Cancellation is Illegal

Counterexample: Let \( a=2, b=3, c=1 \).

Left: \( \dfrac{5}{3} \). "Cancelled" right: \( \dfrac{3}{1} = 3 \). Since \( \dfrac{5}{3} \neq 3 \), the cancellation fails.

Algebraic argument: Cancellation works when a common factor divides the entire numerator: \( \dfrac{a \cdot k}{a \cdot m} = \dfrac{k}{m} \). But \( a+b \) cannot be written as \( a \cdot (\text{polynomial}) \) in general — \( a \) is a term, not a factor, of the numerator.

← Back to Lesson