Section 7: Solving Rational Equations

# Elementary and Intermediate Algebra: Graphs & Models (3rd Edition)

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Section 7.7 Solving Rational Equations 711 Version: Fall 2007 7.7 Solving Rational Equations When simplifying complex fractions in the previous section, we saw that multiplying both numerator and denominator by the appropriate expression could “clear” all frac- tions from the numerator and denominator, greatly simplifying the rational expression. In this section, a similar technique is used. Clear the Fractions from a Rational Equation. If your equation has rational expressions, multiply both sides of the equation by the least common denominator to clear the equation of rational expressions. Let’s look at an example. l⚏ Example 1. Solve the following equation for x . x 2 2 3 = 3 4 (2) To clear this equation of fractions, we will multiply both sides by the common denominator for 2, 3, and 4, which is 12. Distribute 12 in the second step. 12 x 2 2 3 = 3 4 12 12 x 2 12 2 3 = 3 4 12 Multiply. 6 x 8 = 9 We’ve succeeded in clearing the rational expressions from the equation by multiply- ing through by the common denominator. We now have a simple linear equation which can be solved by first adding 8 to both sides of the equation, followed by dividing both sides of the equation by 6. 6 x = 17 x = 17 6 We’ll leave it to our readers to check this solution. Copyrighted material. See: 1

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712 Chapter 7 Rational Functions Version: Fall 2007 Let’s try another example. l⚏ Example 3. Solve the following equation for x . 6 = 5 x + 6 x 2 (4) In this equation, the denominators are 1, x , and x 2 , and the common denominator for both sides of the equation is x 2 . Consequently, we begin the solution by first multiplying both sides of the equation by x 2 . x 2 (6) = 5 x + 6 x 2 x 2 x 2 (6) = 5 x x 2 + 6 x 2 x 2 Simplify. 6 x 2 = 5 x + 6 Note that multiplying both sides of the original equation by the least common denominator clears the equation of all rational expressions. This last equation is non- linear, 2 so make one side of the equation equal to zero by subtracting 5 x and 6 from both sides of the equation. 6 x 2 5 x 6 = 0 To factor the left-hand side of this equation, note that it is a quadratic trinomial with ac = (6)( 6) = 36 . The integer pair 4 and 9 have product 36 and sum 5 . Split the middle term using this pair and factor by grouping. 6 x 2 + 4 x 9 x 6 = 0 2 x (3 x + 2) 3(3 x + 2) = 0 (2 x 3)(3 x + 2) = 0 The zero product property forces either 2 x 3 = 0 or 3 x + 2 = 0 . Each of these linear equations is easily solved. x = 3 2 or x = 2 3 Of course, we should always check our solutions. Substituting x = 3 / 2 into the right-hand side of the original equation (4) , Whenever an equation in x has a power of x other than 1, the equation is nonlinear (the graphs 2 involved are not all lines). As we’ve seen in previous chapters, the approach to solving a quadratic (second degree) equation should be to make one side of the equation equal to zero, then factor or use the quadratic formula to find the solutions.
Section 7.7 Solving Rational Equations 713 Version: Fall 2007 5 x + 6 x 2 = 5 3 / 2 + 6 (3 / 2) 2 = 5 3 / 2 + 6 9 / 4 .

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