Guide to Solving Rational Equations. Step-by-step guided exercises: from determining the domain to clearing the denominators. A comprehensive collection to understand when a solution is valid or must be rejected as extraneous.
Exercise 1 — 20/03/2026 — level ★★☆☆☆
\[ \frac{1}{x} = 3 \]
Answer
\[ x = \dfrac{1}{3} \]
Full 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 — 20/03/2026 — level ★★☆☆☆
\[ \frac{2}{x - 1} = 4 \]
Answer
\[ x = \dfrac{3}{2} \]
Full 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 — 20/03/2026 — level ★★☆☆☆
\[ \frac{x}{x - 3} = 2 \]
Answer
\[ x = 6 \]
Full 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 — 20/03/2026 — level ★★☆☆☆
\[ \frac{x + 1}{x - 2} = 3 \]
Answer
\[ x = \dfrac{7}{2} \]
Full 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 — 20/03/2026 — level ★★★☆☆
\[ \frac{1}{x} + \frac{1}{3} = \frac{5}{6} \]
Answer
\[ x = 2 \]
Full 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 — 20/03/2026 — level ★★★☆☆
\[ \frac{2}{x - 1} - \frac{1}{x + 1} = 0 \]
Answer
\[ x = -3 \]
Full 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 — 20/03/2026 — level ★★★☆☆
\[ \frac{3}{x} + \frac{2}{x + 2} = 2 \]
Answer
\[ x = -\dfrac{3}{2} \quad \text{or} \quad x = 2 \]
Full 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 — 20/03/2026 — level ★★★☆☆
\[ \frac{2}{x} - \frac{3}{x + 1} = \frac{1}{x(x + 1)} \]
Answer
\[ x = 1 \]
Full 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 — 20/03/2026 — level ★★★☆☆
\[ \frac{1}{x - 2} + \frac{1}{x + 2} = \frac{4}{x^2 - 4} \]
Answer
\[ \text{No solution} \]
Full 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 — 20/03/2026 — level ★★★☆☆
\[ \frac{3x - 1}{x + 2} = \frac{x + 1}{x - 2} \]
Answer
\[ x = 0 \quad \text{or} \quad x = 5 \]
Full 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 — 20/03/2026 — level ★★★★☆
\[ \frac{x}{x - 2} + \frac{2}{x + 1} = 1 \]
Answer
\[ x = \dfrac{1}{2} \]
Full 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 — 20/03/2026 — level ★★★★☆
\[ \frac{3}{x + 1} - \frac{2}{x - 2} = \frac{1}{x^2 - x - 2} \]
Answer
\[ x = 9 \]
Full 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 — 20/03/2026 — level ★★★★☆
\[ \frac{2}{x - 1} + \frac{3}{x + 2} = \frac{5}{x^2 + x - 2} \]
Answer
\[ x = \dfrac{4}{5} \]
Full 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 — 20/03/2026 — level ★★★★☆
\[ \frac{x}{x - 2} - \frac{4}{x^2 - 4} = \frac{1}{x + 2} \]
Answer
\[ x = 1 \]
Full 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 — 20/03/2026 — level ★★★★☆
\[ \frac{x + 2}{x - 1} + \frac{x - 2}{x + 1} = 4 \]
Answer
\[ x = 2 \quad \text{or} \quad x = -2 \]
Full 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 — 20/03/2026 — level ★★★★☆
\[ \frac{x + 2}{x + 1} - \frac{x}{x - 1} = \frac{4}{x^2 - 1} \]
Answer
\[ \text{No solution} \]
Full 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 — 20/03/2026 — level ★★★★★
\[ \frac{3}{x - 2} - \frac{2}{x + 1} = \frac{5}{x^2 - x - 2} \]
Answer
\[ x = -2 \]
Full 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 — 20/03/2026 — level ★★★★★
\[ \frac{x}{x - 1} - \frac{1}{x + 2} = \frac{x^2 + 5}{x^2 + x - 2} \]
Answer
\[ x = 4 \]
Full 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 — 20/03/2026 — level ★★★★★
\[ \frac{x + 1}{x - 1} + \frac{x - 1}{x + 1} = \frac{10}{3} \]
Answer
\[ x = 2 \quad \text{or} \quad x = -2 \]
Full 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 — 20/03/2026 — level ★★★★★
\[ \frac{1}{x - 1} + \frac{1}{x + 1} = \frac{2}{x^2 - 1} + 1 \]
Answer
\[ \text{No solution} \]
Full 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
\[ \boxed{\text{Equation has no solution}} \]