A progressive collection of 20 step-by-step exercises to learn how to determine the domain, clear denominators and check whether each solution is valid or must be rejected as extraneous.
Each exercise shows how to solve rational equations carefully, with particular attention to domain restrictions and final verification.
Exercise 1 — level ★★☆☆☆
\[ \frac{1}{x} = 3 \]
Answer
\[ x = \dfrac{1}{3} \]
Solution
Domain Conditions
The denominator cannot be zero: \(x \neq 0\).
Key Idea
Multiply both sides by the denominator \(x\) to clear the fraction.
Multiplying Through by \(x\)
\[ 1 = 3x \implies x = \frac{1}{3} \]
Check
\[ \frac{1}{1/3} = 3 \]
Answer
\[ \boxed{x = \dfrac{1}{3}} \]
Exercise 2 — level ★★☆☆☆
\[ \frac{2}{x - 1} = 4 \]
Answer
\[ x = \dfrac{3}{2} \]
Solution
Domain Conditions
\[ x - 1 \neq 0 \implies x \neq 1 \]
Multiplying Through by \((x-1)\)
\[ 2 = 4(x-1) = 4x - 4 \implies 4x = 6 \implies x = \frac{3}{2} \]
Check
\[ \frac{2}{\frac{3}{2}-1} = \frac{2}{\frac{1}{2}} = 4 \]
Answer
\[ \boxed{x = \dfrac{3}{2}} \]
Exercise 3 — level ★★☆☆☆
\[ \frac{x}{x - 3} = 2 \]
Answer
\[ x = 6 \]
Solution
Domain Conditions
\[ x \neq 3 \]
Multiplying Through by \((x-3)\)
\[ x = 2(x-3) = 2x - 6 \implies -x = -6 \implies x = 6 \]
Check
\[ \frac{6}{6-3} = \frac{6}{3} = 2 \]
Answer
\[ \boxed{x = 6} \]
Exercise 4 — level ★★☆☆☆
\[ \frac{x + 1}{x - 2} = 3 \]
Answer
\[ x = \dfrac{7}{2} \]
Solution
Domain Conditions
\[ x \neq 2 \]
Multiplying Through by \((x-2)\)
\[ x+1 = 3(x-2) = 3x-6 \implies -2x = -7 \implies x = \frac{7}{2} \]
Check
\[ \frac{\frac{7}{2}+1}{\frac{7}{2}-2} = \frac{\frac{9}{2}}{\frac{3}{2}} = 3 \]
Answer
\[ \boxed{x = \dfrac{7}{2}} \]
Exercise 5 — level ★★★☆☆
\[ \frac{1}{x} + \frac{1}{3} = \frac{5}{6} \]
Answer
\[ x = 2 \]
Solution
Domain Conditions
\[ x \neq 0 \]
Multiplying Through by the LCD
The LCD of \(x\), \(3\) and \(6\) is \(6x\). Multiplying everything by \(6x\):
\[ 6 + 2x = 5x \implies 6 = 3x \implies x = 2 \]
Check
\[ \frac{1}{2} + \frac{1}{3} = \frac{3}{6}+\frac{2}{6} = \frac{5}{6} \]
Answer
\[ \boxed{x = 2} \]
Exercise 6 — level ★★★☆☆
\[ \frac{2}{x - 1} - \frac{1}{x + 1} = 0 \]
Answer
\[ x = -3 \]
Solution
Domain Conditions
\[ x \neq 1 \qquad x \neq -1 \]
Multiplying Through by the LCD \((x-1)(x+1)\)
\[ 2(x+1) - (x-1) = 0 \implies 2x+2-x+1 = 0 \implies x+3 = 0 \implies x = -3 \]
Check
\[ \frac{2}{-4} - \frac{1}{-2} = -\frac{1}{2}+\frac{1}{2} = 0 \]
Answer
\[ \boxed{x = -3} \]
Exercise 7 — level ★★★☆☆
\[ \frac{3}{x} + \frac{2}{x + 2} = 2 \]
Answer
\[ x = -\dfrac{3}{2} \quad \text{or} \quad x = 2 \]
Solution
Domain Conditions
\[ x \neq 0 \qquad x \neq -2 \]
Multiplying Through by the LCD \(x(x+2)\)
\[ 3(x+2) + 2x = 2x(x+2) \implies 5x+6 = 2x^2+4x \implies 2x^2-x-6 = 0 \]
Discriminant and Solutions
\[ \Delta = 1+48 = 49 \implies x = \frac{1\pm7}{4} \]
\(x = 2\) or \(x = -\tfrac{3}{2}\). Both satisfy the domain conditions.
Check
\(x=2\): \(\tfrac{3}{2}+\tfrac{2}{4}=\tfrac{3}{2}+\tfrac{1}{2}=2\) \(x=-\tfrac{3}{2}\): \(-2+\tfrac{2}{1/2}=-2+4=2\)
Answer
\[ \boxed{x = -\dfrac{3}{2} \quad \text{or} \quad x = 2} \]
Exercise 8 — level ★★★☆☆
\[ \frac{2}{x} - \frac{3}{x + 1} = \frac{1}{x(x + 1)} \]
Answer
\[ x = 1 \]
Solution
Domain Conditions
\[ x \neq 0 \qquad x \neq -1 \]
Multiplying Through by the LCD \(x(x+1)\)
\[ 2(x+1) - 3x = 1 \implies 2x+2-3x = 1 \implies -x+2 = 1 \implies x = 1 \]
Check
\[ \frac{2}{1}-\frac{3}{2} = 2-\frac{3}{2} = \frac{1}{2} \qquad \frac{1}{1\cdot2} = \frac{1}{2} \]
Answer
\[ \boxed{x = 1} \]
Exercise 9 — level ★★★☆☆
\[ \frac{1}{x - 2} + \frac{1}{x + 2} = \frac{4}{x^2 - 4} \]
Answer
\[ \text{No solution} \]
Solution
Domain Conditions
Since \(x^2-4=(x-2)(x+2)\), the conditions are:
\[ x \neq 2 \qquad x \neq -2 \]
Multiplying Through by the LCD \((x-2)(x+2)\)
\[ (x+2)+(x-2) = 4 \implies 2x = 4 \implies x = 2 \]
Analyzing the Result
The value \(x=2\) is excluded by the domain conditions: it is an extraneous solution introduced by clearing the denominators.
Answer
\[ \boxed{\text{Equation has no solution}} \]
Exercise 10 — level ★★★☆☆
\[ \frac{3x - 1}{x + 2} = \frac{x + 1}{x - 2} \]
Answer
\[ x = 0 \quad \text{or} \quad x = 5 \]
Solution
Domain Conditions
\[ x \neq -2 \qquad x \neq 2 \]
Cross-Multiplication
\[ (3x-1)(x-2) = (x+1)(x+2) \]
\[ 3x^2-7x+2 = x^2+3x+2 \implies 2x^2-10x = 0 \implies 2x(x-5) = 0 \]
Solutions
\(x=0\) or \(x=5\). Both satisfy the domain conditions.
Check
\(x=0\): \(\tfrac{-1}{2}=\tfrac{1}{-2}=-\tfrac{1}{2}\) \(x=5\): \(\tfrac{14}{7}=2\) and \(\tfrac{6}{3}=2\)
Answer
\[ \boxed{x = 0 \quad \text{or} \quad x = 5} \]
Exercise 11 — level ★★★★☆
\[ \frac{x}{x - 2} + \frac{2}{x + 1} = 1 \]
Answer
\[ x = \dfrac{1}{2} \]
Solution
Domain Conditions
\[ x \neq 2 \qquad x \neq -1 \]
Multiplying Through by the LCD \((x-2)(x+1)\)
\[ x(x+1) + 2(x-2) = (x-2)(x+1) \]
\[ x^2+x+2x-4 = x^2-x-2 \]
\[ 3x-4 = -x-2 \implies 4x = 2 \implies x = \frac{1}{2} \]
Check
\[ \frac{1/2}{-3/2}+\frac{2}{3/2} = -\frac{1}{3}+\frac{4}{3} = 1 \]
Answer
\[ \boxed{x = \dfrac{1}{2}} \]
Exercise 12 — level ★★★★☆
\[ \frac{3}{x + 1} - \frac{2}{x - 2} = \frac{1}{x^2 - x - 2} \]
Answer
\[ x = 9 \]
Solution
Domain Conditions
Since \(x^2-x-2=(x+1)(x-2)\), the conditions are:
\[ x \neq -1 \qquad x \neq 2 \]
Multiplying Through by the LCD \((x+1)(x-2)\)
\[ 3(x-2) - 2(x+1) = 1 \implies 3x-6-2x-2 = 1 \implies x-8 = 1 \implies x = 9 \]
Check
\[ \frac{3}{10}-\frac{2}{7} = \frac{21-20}{70} = \frac{1}{70} \qquad \frac{1}{81-9-2} = \frac{1}{70} \]
Answer
\[ \boxed{x = 9} \]
Exercise 13 — level ★★★★☆
\[ \frac{2}{x - 1} + \frac{3}{x + 2} = \frac{5}{x^2 + x - 2} \]
Answer
\[ x = \dfrac{4}{5} \]
Solution
Domain Conditions
Since \(x^2+x-2=(x-1)(x+2)\), the conditions are:
\[ x \neq 1 \qquad x \neq -2 \]
Multiplying Through by the LCD \((x-1)(x+2)\)
\[ 2(x+2) + 3(x-1) = 5 \implies 2x+4+3x-3 = 5 \implies 5x+1 = 5 \implies x = \frac{4}{5} \]
Check
\[ \frac{2}{-1/5}+\frac{3}{14/5} = -10+\frac{15}{14} = \frac{-140+15}{14} = -\frac{125}{14} \]
\[ \frac{5}{\frac{16}{25}+\frac{4}{5}-2} = \frac{5}{-14/25} = -\frac{125}{14} \]
Answer
\[ \boxed{x = \dfrac{4}{5}} \]
Exercise 14 — level ★★★★☆
\[ \frac{x}{x - 2} - \frac{4}{x^2 - 4} = \frac{1}{x + 2} \]
Answer
\[ x = 1 \]
Solution
Domain Conditions
Since \(x^2-4=(x-2)(x+2)\), the conditions are:
\[ x \neq 2 \qquad x \neq -2 \]
Multiplying Through by the LCD \((x-2)(x+2)\)
\[ x(x+2)-4 = (x-2) \]
\[ x^2+2x-4 = x-2 \implies x^2+x-2 = 0 \implies (x+2)(x-1) = 0 \]
Checking and Discarding
\(x=-2\): excluded by the domain conditions, discarded.
\(x=1\): \(\tfrac{1}{-1}-\tfrac{4}{-3}=-1+\tfrac{4}{3}=\tfrac{1}{3}\) and \(\tfrac{1}{3}\) ✓
Answer
\[ \boxed{x = 1} \]
Exercise 15 — level ★★★★☆
\[ \frac{x + 2}{x - 1} + \frac{x - 2}{x + 1} = 4 \]
Answer
\[ x = 2 \quad \text{or} \quad x = -2 \]
Solution
Domain Conditions
\[ x \neq 1 \qquad x \neq -1 \]
Multiplying Through by the LCD \((x-1)(x+1)\)
\[ (x+2)(x+1)+(x-2)(x-1) = 4(x^2-1) \]
\[ (x^2+3x+2)+(x^2-3x+2) = 4x^2-4 \]
\[ 2x^2+4 = 4x^2-4 \implies 2x^2 = 8 \implies x^2 = 4 \implies x = \pm2 \]
Check
\(x=2\): \(\tfrac{4}{1}+\tfrac{0}{3}=4\) \(x=-2\): \(\tfrac{0}{-3}+\tfrac{-4}{-1}=0+4=4\)
Answer
\[ \boxed{x = 2 \quad \text{or} \quad x = -2} \]
Exercise 16 — level ★★★★☆
\[ \frac{x + 2}{x + 1} - \frac{x}{x - 1} = \frac{4}{x^2 - 1} \]
Answer
\[ \text{No solution} \]
Solution
Domain Conditions
Since \(x^2-1=(x+1)(x-1)\), the conditions are:
\[ x \neq -1 \qquad x \neq 1 \]
Multiplying Through by the LCD \((x+1)(x-1)\)
\[ (x+2)(x-1) - x(x+1) = 4 \]
\[ (x^2+x-2)-(x^2+x) = 4 \implies -2 = 4 \]
Analyzing the Result
A numerical contradiction arises: the equation is inconsistent for every value of \(x\) and has no solution.
Answer
\[ \boxed{\text{Inconsistent equation — no solution}} \]
Exercise 17 — level ★★★★★
\[ \frac{3}{x - 2} - \frac{2}{x + 1} = \frac{5}{x^2 - x - 2} \]
Answer
\[ x = -2 \]
Solution
Domain Conditions
Since \(x^2-x-2=(x-2)(x+1)\), the conditions are:
\[ x \neq 2 \qquad x \neq -1 \]
Multiplying Through by the LCD \((x-2)(x+1)\)
\[ 3(x+1)-2(x-2) = 5 \implies 3x+3-2x+4 = 5 \implies x+7 = 5 \implies x = -2 \]
Check
\[ \frac{3}{-4}-\frac{2}{-1} = -\frac{3}{4}+2 = \frac{5}{4} \qquad \frac{5}{4+2-2} = \frac{5}{4} \]
Answer
\[ \boxed{x = -2} \]
Exercise 18 — level ★★★★★
\[ \frac{x}{x - 1} - \frac{1}{x + 2} = \frac{x^2 + 5}{x^2 + x - 2} \]
Answer
\[ x = 4 \]
Solution
Domain Conditions
Since \(x^2+x-2=(x-1)(x+2)\), the conditions are:
\[ x \neq 1 \qquad x \neq -2 \]
Multiplying Through by the LCD \((x-1)(x+2)\)
\[ x(x+2)-(x-1) = x^2+5 \]
\[ x^2+2x-x+1 = x^2+5 \implies x+1 = 5 \implies x = 4 \]
Check
\[ \frac{4}{3}-\frac{1}{6} = \frac{8}{6}-\frac{1}{6} = \frac{7}{6} \qquad \frac{21}{18} = \frac{7}{6} \]
Answer
\[ \boxed{x = 4} \]
Exercise 19 — level ★★★★★
\[ \frac{x + 1}{x - 1} + \frac{x - 1}{x + 1} = \frac{10}{3} \]
Answer
\[ x = 2 \quad \text{or} \quad x = -2 \]
Solution
Domain Conditions
\[ x \neq 1 \qquad x \neq -1 \]
Substitution \(t = \dfrac{x+1}{x-1}\)
We notice that \(\dfrac{x-1}{x+1} = \dfrac{1}{t}\). The equation becomes:
\[ t + \frac{1}{t} = \frac{10}{3} \implies 3t^2-10t+3 = 0 \implies (3t-1)(t-3) = 0 \]
Case \(t = 3\)
\[ \frac{x+1}{x-1} = 3 \implies x+1 = 3x-3 \implies x = 2 \]
Case \(t = \frac{1}{3}\)
\[ \frac{x+1}{x-1} = \frac{1}{3} \implies 3x+3 = x-1 \implies x = -2 \]
Check
\(x=2\): \(3+\tfrac{1}{3}=\tfrac{10}{3}\) \(x=-2\): \(\tfrac{-1}{-3}+\tfrac{-3}{-1}=\tfrac{1}{3}+3=\tfrac{10}{3}\)
Answer
\[ \boxed{x = 2 \quad \text{or} \quad x = -2} \]
Exercise 20 — level ★★★★★
\[ \frac{1}{x - 1} + \frac{1}{x + 1} = \frac{2}{x^2 - 1} + 1 \]
Answer
\[ \text{No solution} \]
Solution
Domain Conditions
Since \(x^2-1=(x-1)(x+1)\), the conditions are:
\[ x \neq 1 \qquad x \neq -1 \]
Multiplying Through by the LCD \((x-1)(x+1)\)
\[ (x+1)+(x-1) = 2+(x^2-1) \implies 2x = x^2+1 \implies x^2-2x+1 = 0 \implies (x-1)^2 = 0 \]
Analyzing the Result
The only algebraic solution would be \(x=1\), but it is excluded by the domain conditions. This is an extraneous solution: the equation has no valid solution.
Answer
\[ \boxed{\text{Equation has no solution}} \]