To solve a rational equation, we will multiply both sides by the LCD of all rational expressions. We must be careful since multiplying both sides by a variable or variable expression will sometimes give us extraneous solutions. Therefore, we must always check our proposed solutions for a rational equation in the original equation.

Test Objectives
• Demonstrate an understanding of how to find the domain for a rational expression
• Demonstrate the ability to solve a rational equation
Solving Rational Equations Practice Test:

#1:

Instructions: solve each equation.

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$$a)\hspace{.2em}\frac{1}{x^2 + 10x + 25}=\frac{1}{x + 5}- \frac{2}{x^2 + 10x + 25}$$

$$b)\hspace{.2em}\frac{1}{x + 1}=\frac{5}{x + 1}+ \frac{x + 7}{x^2 + x}$$

#2:

Instructions: solve each equation.

$$a)\hspace{.2em}\frac{x + 2}{x^2 - 3x}+ \frac{7}{x}=\frac{1}{x - 3}$$

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$$b)\hspace{.2em}\frac{4}{5x + 7}-\frac{6}{5x^3 + 52x^2 + 63x}=\frac{x + 3}{5x^3 + 52x^2 + 63x}$$

#3:

Instructions: solve each equation.

$$a)\hspace{.2em}\frac{3x - 24}{x^2 - 6x}=\frac{1}{x - 6}+ \frac{1}{x}$$

$$b)\hspace{.2em}1=\frac{1}{x}+ \frac{x^2 - 2x - 3}{x^2 + 6x}$$

#4:

Instructions: solve each equation.

$$a)\hspace{.2em}\frac{x}{2x^2 + 10x + 8}- \frac{x - 7}{x + 4}=\frac{1}{x + 1}$$

$$b)\hspace{.2em}1 - \frac{7}{x}=\frac{6x^2 + 12x - 18}{x^2 + 4x}$$

#5:

Instructions: solve each equation.

$$a)\hspace{.2em}1 - \frac{2x - 6}{x}=\frac{3x^2 + 21x - 24}{x^2 - 5x}$$

$$b)\hspace{.2em}\frac{x + 5}{x - 4}+ \frac{1}{x - 4}=\frac{7}{x^2 - 4x}$$

Written Solutions:

#1:

Solutions:

$$a)\hspace{.2em}x=-2$$

$$b)\hspace{.2em}x=-\frac{7}{5}$$

#2:

Solutions:

$$a)\hspace{.2em}x=\frac{19}{7}$$

$$b)\hspace{.2em}x=\frac{1}{4}$$

#3:

Solutions:

$$a)\hspace{.2em}x=18$$

$$b)\hspace{.2em}x=\frac{3}{7}$$

#4:

Solutions:

$$a)\hspace{.2em}x=-\frac{1}{2}, 6$$

$$b)\hspace{.2em}x=-2,-1$$

#5:

Solutions:

$$a)\hspace{.2em}x=-\frac{3}{2}, -1$$

$$b)\hspace{.2em}x=-7, 1$$