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

## How to Solve Rational Equations

At the start of our Algebra 2 course, we learned how to solve a linear equation in one variable with fractions involved. To solve a linear equation in one variable with fractions involved, we multiply both sides of the equation by the LCD of all fractions. This process will clear the equation of fractions. Suppose we had the following equation: $$2x + \frac{1}{3}x=-\frac{8}{9}+ \frac{5}{3}x$$ We have three fractions in our equation: 1/3, -8/9, and 5/3. The LCD of all fractions is 9. We will clear our fractions by multiplying both sides of the equation by 9: $$9\left(2x + \frac{1}{3}x\right)=9\left(-\frac{8}{9}+ \frac{5}{3}x\right)$$ $$18x + 3x=-8 + 15x$$ $$x=-\frac{4}{3}$$

### Solving Rational Equations

An equation with rational expressions involved is known as a "rational equation". We can solve rational equations by multiplying both sides of the equation by the LCD of all rational expressions. There is one problem that comes up. When we multiply both sides of an equation by a variable expression, the resulting solutions may not satisfy the original equation. Recall that a rational expression is only defined for values that do not result in a denominator of zero. When one of our proposed solutions results in a denominator of zero, we must reject the solution. These types of solutions are known as "extraneous solutions" and do not satisfy the original equation. Let's look at some examples.
Example 1: Solve each equation $$\frac{1}{x}=\frac{8}{x^2 - 4x}- \frac{1}{x - 4}$$ Step 1) Find the LCD
Let's begin by factoring the denominators, this will allow us to find the LCD. $$\frac{1}{x}=\frac{8}{x(x - 4)}- \frac{1}{x - 4}$$ LCD: $$x(x - 4)$$ Step 2) Multiply each side of the equation by the LCD: $$x(x - 4) \cdot \frac{1}{x}=x(x - 4)\left(\frac{8}{x(x - 4)}- \frac{1}{(x - 4)}\right)$$ Left side: $$x(x - 4) \cdot \frac{1}{x}$$ $$\require{cancel}\cancel{x}(x - 4) \cdot \frac{1}{\cancel{x}}\hspace{.2em}» \hspace{.2em}x - 4$$ Right side: $$x(x - 4)\left(\frac{8}{x(x - 4)}- \frac{1}{(x - 4)}\right)$$ $$x(x - 4) \cdot \frac{8}{x(x - 4)}- x(x-4) \cdot \frac{1}{(x - 4)}$$ $$\cancel{x(x - 4)}\cdot \frac{8}{\cancel{x(x - 4)}}- x\cancel{(x-4)}\cdot \frac{1}{\cancel{(x - 4)}}\hspace{.2em}» \hspace{.2em}8 - x$$ Step 3) Solve the equation: $$x - 4=8 - x$$ $$2x=12$$ $$x=6$$ Step 4) Check the solution in the original equation. $$\frac{1}{x}=\frac{8}{x^2 - 4x}- \frac{1}{x - 4}$$ $$\frac{1}{6}=\frac{8}{(6)^2 - 4(6)}- \frac{1}{(6) - 4}$$ $$\frac{1}{6}=\frac{8}{12}- \frac{1}{2}$$ $$\require{color}\frac{1}{6}=\frac{1}{6}\hspace{.2em}\color{green}{✓}$$ Example 2: Solve each equation $$\frac{1}{x-4}+ \frac{1}{6x^2 - 29x + 20}=\frac{4x + 16}{6x^2 - 29x + 20}$$ Step 1) Find the LCD
Let's begin by factoring the denominators, this will allow us to find the LCD. $$\frac{1}{(x-4)}+ \frac{1}{(x - 4)(6x - 5)}=\frac{4x + 16}{(x - 4)(6x - 5)}$$ LCD: $$(x - 4)(6x - 5)$$ Step 2) Multiply each side of the equation by the LCD: $$(x - 4)(6x - 5) \cdot \left(\frac{1}{(x-4)}+ \frac{1}{(x - 4)(6x - 5)}\right)=(x - 4)(6x - 5) \cdot \frac{4x + 16}{(x - 4)(6x - 5)}$$ Left Side: $$(x - 4)(6x - 5) \cdot \frac{1}{(x-4)}+ (x - 4)(6x - 5) \cdot \frac{1}{(x - 4)(6x - 5)}$$ $$\cancel{(x - 4)}(6x - 5) \cdot \frac{1}{\cancel{(x-4)}}+ \cancel{(x - 4)(6x - 5)}\cdot \frac{1}{\cancel{(x - 4)(6x - 5)}}\hspace{.2em}» \hspace{.2em}6x - 4$$ Right Side: $$(x - 4)(6x - 5) \cdot \frac{4x + 16}{(x - 4)(6x - 5)}$$ $$\cancel{(x - 4)(6x - 5)}\cdot \frac{4x + 16}{\cancel{(x - 4)(6x - 5)}}\hspace{.2em}» \hspace{.2em}4x + 16$$ Step 3) Solve the equation: $$6x - 4=4x + 16$$ $$2x=20$$ $$x=10$$ Step 4) Check the solution in the original equation. $$\frac{1}{x-4}+ \frac{1}{6x^2 - 29x + 20}=\frac{4x + 16}{6x^2 - 29x + 20}$$ $$\frac{1}{(10)-4}+ \frac{1}{6(10)^2 - 29(10) + 20}=\frac{4(10) + 16}{6(10)^2 - 29(10) + 20}$$ $$\frac{28}{165}=\frac{28}{165}\hspace{.2em}\color{green}{✓}$$ Example 3: Solve each equation $$\frac{1}{x^3 - 12x^2 + 36x}=\frac{x + 1}{x^2 - 6x}+ \frac{7}{x^3 - 12x^2 + 36x}$$ Step 1) Find the LCD
Let's begin by factoring the denominators, this will allow us to find the LCD. $$\frac{1}{x(x - 6)(x - 6)}=\frac{x + 1}{x(x - 6)}+ \frac{7}{x(x - 6)(x - 6)}$$ LCD: $$x(x - 6)(x - 6)$$ Step 2) Multiply each side of the equation by the LCD: $$x(x - 6)(x - 6) \cdot \frac{1}{x(x - 6)(x - 6)}=x(x - 6)(x - 6)\left(\frac{x + 1}{x(x - 6)}+ \frac{7}{x(x - 6)(x - 6)}\right)$$ Left Side: $$x(x - 6)(x - 6) \cdot \frac{1}{x(x - 6)(x - 6)}$$ $$\cancel{x(x - 6)(x - 6)}\cdot \frac{1}{\cancel{x(x - 6)(x - 6)}}\hspace{.2em}» \hspace{.2em}1$$ Right Side: $$x(x - 6)(x - 6) \cdot \frac{x + 1}{x(x - 6)}+ x(x - 6)(x - 6) \cdot \frac{7}{x(x - 6)(x - 6)}$$ $$\cancel{x(x - 6)}(x - 6) \cdot \frac{x + 1}{\cancel{x(x - 6)}}+ \cancel{x(x - 6)(x - 6)}\cdot \frac{7}{\cancel{x(x - 6)(x - 6)}}\hspace{.2em}» \hspace{.2em}(x-6)(x + 1) + 7$$ Step 3) Solve the Equation: $$1=(x - 6)(x + 1) + 7$$ $$1=x^2 - 5x - 6 + 7$$ $$1=x^2 - 5x + 1$$ $$x^2 - 5x=0$$ $$x(x - 5)=0$$ $$x=0$$ $$x - 5=0$$ $$x=5$$ Our proposed solutions are x = 0, and x = 5.
Step 4) Check the solutions in the original equation. $$\frac{1}{x^3 - 12x^2 + 36x}=\frac{x + 1}{x^2 - 6x}+ \frac{7}{x^3 - 12x^2 + 36x}$$ $$\frac{1}{(0)^3 - 12(0)^2 + 36(0)}=\frac{(0) + 1}{(0)^2 - 6(0)}+ \frac{7}{(0)^3 - 12(0)^2 + 36(0)}$$ $$\frac{1}{0}=\frac{1}{0}+ \frac{7}{0}$$ We will reject the extraneous solution x = 0 since it results in division by zero. Let's check our other proposed solution of x = 5. $$\frac{1}{x^3 - 12x^2 + 36x}=\frac{x + 1}{x^2 - 6x}+ \frac{7}{x^3 - 12x^2 + 36x}$$ $$\frac{1}{(5)^3 - 12(5)^2 + 36(5)}=\frac{(5) + 1}{(5)^2 - 6(5)}+ \frac{7}{(5)^3 - 12(5)^2 + 36(5)}$$ $$\frac{1}{5}=\frac{-6}{5}+ \frac{7}{5}$$ $$\frac{1}{5}=\frac{1}{5}\hspace{.2em}\color{green}{✓}$$ The only solution to this equation is x = 5.