Equations with Rational Expressions - Step by Step Tutorial

This tutorial explains how to solve equations that contain rational expressions. The main strategy is to multiply all terms of the equation by the lowest common denominator (LCD) to eliminate fractions.

Examples with Solutions

Example 1

Solve the following equation:

\[ \frac{10}{x - 1} = 5 \]

View Solution
  1. Multiply both sides by the denominator \(x - 1\) to eliminate the fraction: \[ (x - 1)\left(\frac{10}{x - 1}\right) = 5(x - 1) \]
  2. Simplify each side: \[ 10 = 5x - 5 \]
  3. Add 5 to both sides to isolate the variable term: \[ 15 = 5x \]
  4. Divide by 5: \[ x = 3 \]
  5. Check the solution:

    Substituting \(x = 3\) back into the original equation:

    \[ \frac{10}{3 - 1} = \frac{10}{2} = 5 \]

    Valid solution.

Conclusion: The solution is \[ x = 3 \]

Example 2

Solve the following equation:

\[ 1 - \frac{1}{x - 2} = 4 \]

View Solution
  1. Multiply all terms by the denominator \(x - 2\) to eliminate the fraction: \[ (x - 2)\left(1 - \frac{1}{x - 2}\right) = (x - 2) \cdot 4 \]
  2. Simplify each side: \[ (x - 2) - 1 = 4(x - 2) \]
  3. Expand and simplify: \[ x - 3 = 4x - 8 \]
  4. Move all terms containing \(x\) to one side: \[ -3x - 3 = -8 \]
  5. Isolate \(x\): \[ -3x = -5 \] \[ x = \frac{5}{3} \]

Conclusion: The solution is \[ x = \frac{5}{3} \]

Example 3

Find all real solutions to the equation:

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

View Solution
  1. Identify the lowest common denominator (LCD): \((x-2)(x+2)\). Multiply all terms by the LCD: \[ (x-2)(x+2)\left(1 - \frac{1}{x - 2}\right) = (x-2)(x+2)\left(-\frac{4}{x^2 - 4}\right) \]
  2. Cancel common factors: \[ (x-2)(x+2) - (x+2) = -4 \]
  3. Expand and simplify: \[ x^2 - x - 6 = -4 \]
  4. Add 4 to both sides: \[ x^2 - x - 2 = 0 \]
  5. Factor the quadratic: \[ (x + 1)(x - 2) = 0 \]
  6. Set each factor to zero and solve: \[ x_1 = -1, \quad x_2 = 2 \]
  7. Check each solution for extraneous roots:

    Check \(x = -1\):

    \[ \text{Left Side} = 1 - \frac{1}{-1 - 2} = 1 - \frac{1}{-3} = \frac{4}{3} \] \[ \text{Right Side} = -\frac{4}{(-1)^2 - 4} = -\frac{4}{1-4} = \frac{4}{3} \]

    Valid solution.

    Check \(x = 2\):

    This value makes the denominator \(x-2=0\), which is undefined. Therefore, it is not a valid solution.

Conclusion: The solution is \[ x = -1 \]

Example 4

Find all real solutions to the equation:

\[ \frac{x}{x - 3} - \frac{2}{x + 3} = \frac{18}{x^2 - 9} \]

View Solution
  1. Identify the lowest common denominator (LCD). Notice that \(x^2 - 9 = (x - 3)(x + 3)\). The LCD is \((x - 3)(x + 3)\).
  2. Multiply all terms by the LCD: \[ (x - 3)(x + 3)\left(\frac{x}{x - 3} - \frac{2}{x + 3}\right) = (x - 3)(x + 3)\left(\frac{18}{x^2 - 9}\right) \]
  3. Cancel common factors in the denominators: \[ x(x + 3) - 2(x - 3) = 18 \]
  4. Expand and simplify the left side: \[ x^2 + 3x - 2x + 6 = 18 \] \[ x^2 + x + 6 = 18 \]
  5. Move all terms to one side to form a standard quadratic equation: \[ x^2 + x - 12 = 0 \]
  6. Factor the quadratic: \[ (x + 4)(x - 3) = 0 \]
  7. Set each factor to zero and solve: \[ x_1 = -4, \quad x_2 = 3 \]
  8. Check each solution for extraneous roots:

    Check \(x = -4\):

    Neither \(x - 3\) nor \(x + 3\) results in zero. The denominators are \(-7\) and \(-1\). Valid solution.

    Check \(x = 3\):

    This value makes the denominator \(x - 3 = 0\), resulting in division by zero. It is an extraneous root and not a valid solution.

Conclusion: The only valid solution is \[ x = -4 \]

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