Lesson Objectives
  • Demonstrate an understanding of how to solve a linear equation with fractions
  • Demonstrate an understanding of how to find the LCD for a group of rational expressions
  • Learn how to solve rational equations (equations with rational expressions)
  • Learn how to reject extraneous solutions for a rational equation

How to Solve Rational Equations


At the beginning of Algebra 1, we learned how to solve equations with fractions. We can clear the fractions from an equation by multiplying both sides of the equation by the LCD of all fractions. Let's take a look at an example: $$-\frac{7}{3}x + \frac{31}{6} = -\frac{1}{2}x + \frac{3}{2}$$ The LCD of the fractions is 6. Let's multiply both sides of the equation by 6 and clear all fractions: $$6 \left(-\frac{7}{3}x + \frac{31}{6}\right) = 6 \left(-\frac{1}{2}x + \frac{3}{2} \right)$$ $$6 \left(-\frac{7}{3}x + \frac{31}{6}\right) = 6 \left(-\frac{1}{2}x + \frac{3}{2} \right)$$ $$6 \left(-\frac{7}{3}x + \frac{31}{6}\right) = $$$$6 \left(-\frac{1}{2}x + \frac{3}{2} \right)$$ Let's distribute our 6 to each term inside of parentheses: $$\require{cancel}6 \cdot -\frac{7}{3}x = 2\cancel{6} \cdot -\frac{7}{\cancel{3}}x = -14x$$ $$6 \cdot \frac{31}{6} = \cancel{6} \cdot \frac{31}{\cancel{6}} = 31$$ $$6 \cdot -\frac{1}{2}x = 3\cancel{6} \cdot -\frac{1}{\cancel{2}}x = -3x$$ $$6 \cdot \frac{3}{2}x = 3\cancel{6} \cdot \frac{3}{\cancel{2}} = 9$$ We now have an equation which is free of all fractions: $$-14x + 31 = -3x + 9$$ $$-14x + 3x = 9 - 31$$ $$-11x = -22$$ $$x = 2$$

Solving Rational Equations

When we solve rational equations, we use a similar process:
  • Multiply each side of the equation by the LCD of all rational expressions to clear the denominators
  • Solve the equation
  • Check each proposed solution by substituting back into the original equation
    • We reject any solution that results in a denominator of zero
It is very important to follow the third step. If we get a solution that results in a denominator of zero, it must be rejected. These solutions are known as extraneous solutions. Let's look at a few examples.
Example 1: Solve each equation $$\frac{5}{3x} = \frac{x + 3}{x^2} + \frac{1}{3x^2}$$ Step 1) Multiply each side of the equation by the LCD of all rational expressions
LCD » 3x2 $$3x^2 \cdot \left(\frac{5}{3x} \right) = 3x^2 \cdot \left(\frac{x + 3}{x^2} + \frac{1}{3x^2} \right)$$ $$3x^2 \cdot \left(\frac{5}{3x} \right) = 3x^2 \cdot \left(\frac{x + 3}{x^2} + \frac{1}{3x^2} \right)$$ $$3x^2 \cdot \left(\frac{5}{3x} \right) = $$$$3x^2 \cdot \left(\frac{x + 3}{x^2} + \frac{1}{3x^2} \right)$$ $$3x^2 \cdot \frac{5}{3x} = \cancel{3x} \cdot x \cdot \frac{5}{\cancel{3x}} = 5x$$ $$3x^2 \cdot \frac{5}{3x} = \cancel{3x} \cdot x \cdot \frac{5}{\cancel{3x}} = 5x$$ $$3x^2 \cdot \frac{5}{3x} = $$$$\cancel{3x} \cdot x \cdot \frac{5}{\cancel{3x}} = 5x$$ $$3x^2 \cdot \frac{x + 3}{x^2} = 3\cancel{x^2} \cdot \frac{x + 3}{\cancel{x^2}} = 3(x + 3)$$ $$3x^2 \cdot \frac{x + 3}{x^2} = 3\cancel{x^2} \cdot \frac{x + 3}{\cancel{x^2}} = 3(x + 3)$$ $$3x^2 \cdot \frac{x + 3}{x^2} = $$$$3\cancel{x^2} \cdot \frac{x + 3}{\cancel{x^2}} = 3(x + 3)$$ $$3x^2 \cdot \frac{1}{3x^2} = \cancel{3x^2} \cdot \frac{1}{\cancel{3x^2}} = 1$$ $$3x^2 \cdot \frac{1}{3x^2} = \cancel{3x^2} \cdot \frac{1}{\cancel{3x^2}} = 1$$ $$3x^2 \cdot \frac{1}{3x^2} = $$$$\cancel{3x^2} \cdot \frac{1}{\cancel{3x^2}} = 1$$ Our equation becomes: $$5x = 3(x + 3) + 1$$ Step 2) Solve the equation: $$5x = 3x + 9 + 1$$ $$2x = 10$$ $$x = 5$$ Step 3) Check the proposed solution of x = 5 $$\frac{5}{3x} = \frac{x + 3}{x^2} + \frac{1}{3x^2}$$ $$\frac{5}{3(5)} = \frac{(5) + 3}{(5)^2} + \frac{1}{3(5)^2}$$ $$\frac{1}{3} = \frac{8}{25} + \frac{1}{75}$$ $$\frac{1}{3} = \frac{24}{75} + \frac{1}{75}$$ $$\frac{1}{3} = \frac{25}{75}$$ $$\frac{1}{3} = \frac{1}{3}$$ Our solution x = 5 is valid.
Example 2: Solve each equation $$\frac{-2}{x + 5} + \frac{3}{x - 5} = \frac{20}{x^2 - 25}$$ Step 1) Multiply each side of the equation by the LCD of all rational expressions
LCD » (x + 5)(x - 5) $$(x + 5)(x - 5) \cdot \left(\frac{-2}{x + 5} + \frac{3}{x - 5} \right) = (x + 5)(x - 5) \cdot \left(\frac{20}{(x + 5)(x - 5)} \right)$$ $$(x + 5)(x - 5) \cdot \left(\frac{-2}{x + 5} + \frac{3}{x - 5} \right) = $$$$(x + 5)(x - 5) \cdot \left(\frac{20}{(x + 5)(x - 5)} \right)$$ $$(x + 5)(x - 5) \hspace{.2em} \cdot $$$$\left(\frac{-2}{x + 5} + \frac{3}{x - 5} \right) = $$$$(x + 5)(x - 5) \hspace{.2em} \cdot $$$$\left(\frac{20}{(x + 5)(x - 5)} \right)$$ $$(x + 5)(x - 5) \cdot \frac{-2}{(x + 5)} = \cancel{(x + 5)}(x - 5) \cdot \frac{-2}{\cancel{(x + 5)}} = -2(x - 5)$$ $$(x + 5)(x - 5) \cdot \frac{-2}{(x + 5)} = $$$$\cancel{(x + 5)}(x - 5) \cdot \frac{-2}{\cancel{(x + 5)}} = -2(x - 5)$$ $$(x + 5)(x - 5) \hspace{.2em} \cdot $$$$\frac{-2}{(x + 5)} = $$$$\cancel{(x + 5)}(x - 5) \hspace{.2em} \cdot $$$$\frac{-2}{\cancel{(x + 5)}} = -2(x - 5)$$ $$(x + 5)(x - 5) \cdot \frac{3}{(x - 5)} = (x + 5)\cancel{(x - 5)} \cdot \frac{3}{\cancel{(x - 5)}} = 3(x + 5)$$ $$(x + 5)(x - 5) \cdot \frac{3}{(x - 5)} = $$$$(x + 5)\cancel{(x - 5)} \cdot \frac{3}{\cancel{(x - 5)}} = 3(x + 5)$$ $$(x + 5)(x - 5) \hspace{.2em} \cdot $$$$\frac{3}{(x - 5)} = $$$$(x + 5)\cancel{(x - 5)} \hspace{.2em} \cdot $$$$\frac{3}{\cancel{(x - 5)}} = 3(x + 5)$$ $$(x + 5)(x - 5) \cdot \frac{20}{(x + 5)(x - 5)} = \cancel{(x + 5)(x - 5)} \cdot \frac{20}{\cancel{(x + 5)(x - 5)}} = 20$$ $$(x + 5)(x - 5) \cdot \frac{20}{(x + 5)(x - 5)} = $$$$\cancel{(x + 5)(x - 5)} \cdot \frac{20}{\cancel{(x + 5)(x - 5)}} = 20$$ $$(x + 5)(x - 5) \hspace{.2em} \cdot $$$$\frac{20}{(x + 5)(x - 5)} = $$$$\cancel{(x + 5)(x - 5)} \hspace{.2em} \cdot $$$$\frac{20}{\cancel{(x + 5)(x - 5)}} = 20$$ Our equation becomes: $$-2(x - 5) + 3(x + 5) = 20$$ Step 2) Solve the equation: $$-2(x - 5) + 3(x + 5) = 20$$ $$-2x + 10 + 3x + 15 = 20$$ $$x + 25 = 20$$ $$x = -5$$ Step 3) Check the proposed solution of x = -5 $$\frac{-2}{x + 5} + \frac{3}{x - 5} = \frac{20}{x^2 - 25}$$ $$\frac{-2}{(-5) + 5} + \frac{3}{(-5) - 5} = \frac{20}{(-5)^2 - 25}$$ $$\frac{-2}{(-5) + 5} + \frac{3}{(-5) - 5} = \frac{20}{(-5)^2 - 25}$$ $$\frac{-2}{(-5) + 5} + \frac{3}{(-5) - 5} = $$$$\frac{20}{(-5)^2 - 25}$$ $$\frac{2}{0} + \frac{3}{-10} = \frac{20}{0}$$ Stop - We can't divide by zero. We must reject our solution of x = -5.
Since there are no other proposed solutions, we can say this equation has no solution.
Example 3: Solve each equation $$\frac{x + 5}{x + 7} + \frac{1}{x + 1} = \frac{x + 3}{x^2 + 8x + 7}$$ Step 1) Multiply each side of the equation by the LCD of all rational expressions
LCD » (x + 7)(x + 1) $$(x + 7)(x + 1) \left(\frac{x + 5}{x + 7} + \frac{1}{x + 1} \right) = (x + 7)(x + 1) \left(\frac{x + 3}{(x + 7)(x + 1)} \right)$$ $$(x + 7)(x + 1) \left(\frac{x + 5}{x + 7} + \frac{1}{x + 1} \right) = $$$$(x + 7)(x + 1) \left(\frac{x + 3}{(x + 7)(x + 1)} \right)$$ $$(x + 7)(x + 1) \left(\frac{x + 5}{x + 7} + \frac{1}{x + 1} \right)$$$$ = $$$$(x + 7)(x + 1) \left(\frac{x + 3}{(x + 7)(x + 1)} \right)$$ $$(x + 7)(x + 1) \cdot \frac{x + 5}{x + 7} = \cancel{(x + 7)}(x + 1) \cdot \frac{x + 5}{\cancel{(x + 7)}} = (x + 1)(x + 5)$$ $$(x + 7)(x + 1) \cdot \frac{x + 5}{x + 7} = $$$$\cancel{(x + 7)}(x + 1) \cdot \frac{x + 5}{\cancel{(x + 7)}} = (x + 1)(x + 5)$$ $$(x + 7)(x + 1) \cdot \frac{x + 5}{x + 7} = $$$$\cancel{(x + 7)}(x + 1) \hspace{.2em} \cdot $$$$\frac{x + 5}{\cancel{(x + 7)}} = (x + 1)(x + 5)$$ $$(x + 7)(x + 1) \cdot \frac{1}{x + 1} = (x + 7)\cancel{(x + 1)} \cdot \frac{1}{\cancel{(x + 1)}} = (x + 7)$$ $$(x + 7)(x + 1) \cdot \frac{1}{x + 1} = $$$$(x + 7)\cancel{(x + 1)} \cdot \frac{1}{\cancel{(x + 1)}} = (x + 7)$$ $$(x + 7)(x + 1) \cdot \frac{1}{x + 1} = $$$$(x + 7)\cancel{(x + 1)} \cdot \frac{1}{\cancel{(x + 1)}}$$$$ = (x + 7)$$ $$(x + 7)(x + 1) \cdot \frac{x + 3}{(x + 7)(x + 1)} = \cancel{(x + 7)(x + 1)} \cdot \frac{x + 3}{\cancel{(x + 7)(x + 1)}} = (x + 3)$$ $$(x + 7)(x + 1) \cdot \frac{x + 3}{(x + 7)(x + 1)} = $$$$\cancel{(x + 7)(x + 1)} \cdot \frac{x + 3}{\cancel{(x + 7)(x + 1)}} = (x + 3)$$ $$(x + 7)(x + 1) \cdot \frac{x + 3}{(x + 7)(x + 1)}$$$$ = $$$$\cancel{(x + 7)(x + 1)} \cdot \frac{x + 3}{\cancel{(x + 7)(x + 1)}}$$$$ = (x + 3)$$ Our equation becomes: $$(x + 1)(x + 5) + (x + 7) = (x + 3)$$ $$(x + 1)(x + 5) + (x + 7) = (x + 3)$$ $$(x + 1)(x + 5) + (x + 7) = $$$$(x + 3)$$ Step 2) Solve the equation: $$(x + 1)(x + 5) + (x + 7) = (x + 3)$$ $$(x + 1)(x + 5) + (x + 7) = (x + 3)$$ $$(x + 1)(x + 5) + (x + 7) = $$$$(x + 3)$$ $$x^2 + 6x + 5 + x + 7 = x + 3$$ $$x^2 + 7x + 12 = x + 3$$ $$x^2 + 6x + 9 = 0$$ $$(x + 3)(x + 3) = 0$$ $$x + 3 = 0$$ $$x = -3$$ Step 3) Check the proposed solution of x = -3 $$\frac{x + 5}{x + 7} + \frac{1}{x + 1} = \frac{x + 3}{x^2 + 8x + 7}$$ $$\frac{(-3) + 5}{(-3) + 7} + \frac{1}{(-3) + 1} = \frac{(-3) + 3}{(-3)^2 + 8(-3) + 7}$$ $$\frac{(-3) + 5}{(-3) + 7} + \frac{1}{(-3) + 1} = \frac{(-3) + 3}{(-3)^2 + 8(-3) + 7}$$ $$\frac{(-3) + 5}{(-3) + 7} + \frac{1}{(-3) + 1} = $$$$\frac{(-3) + 3}{(-3)^2 + 8(-3) + 7}$$ $$\frac{2}{4} + \frac{1}{-2} = \frac{0}{9 - 24 + 7}$$ $$\frac{1}{2} - \frac{1}{2} = 0$$ $$0 = 0$$ Our solution x = -3 is valid.