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By Margaret Lial, Ray Greenwell, Nathan Ritchey

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EXAMPLE 6 Solving Rational Equations Solve each equation. (a) r 2 3r 1 2 5 2 10 15 20 5 SOLUTION The denominators are 10, 15, 20, and 5. Each of these numbers can be divided into 60, so 60 is a common denominator. Multiply both sides of the equation by 60 and use the distributive property. ” Multiply by the common denominator. 4 Equations R-15 Add 29r and 8 to both sides. 6r 2 8 1 1 29r 2 1 8 5 9r 2 12 1 1 29r 2 1 8 23r 5 24 4 r5 3 Multiply each side by 2 13 . Check by substituting into the original equation.

To factor means to factor completely, so that each polynomial factor is prime. 2 EXERCISES Factor each polynomial. If a polynomial cannot be factored, write prime. Factor out the greatest common factor as necessary. 3. 13p4q2 2 39p3q 1 26p2q2 4. 60m4 2 120m3n 1 50m2n2 3 2 5. m2 2 5m 2 14 3 2 6. x2 1 4x 2 5 1. 7a 1 14a 2. 3y 1 24y 1 9y R-8 CHAPTER R Algebra Reference 7. z2 1 9z 1 20 18. 4a2 1 10a 1 6 8. b2 2 8b 1 7 19. 24a4 1 10a3b 2 4a2b2 9. a2 2 6ab 1 5b2 20. 24x4 1 36x3y 2 60x2y2 10. s2 1 2st 2 35t2 2 21.

X x23 x1x 2 32 R-16 CHAPTER R Algebra Reference SOLUTION The common denominator is x 1 x 2 3 2 . Multiply both sides by x 1 x 2 3 2 and solve the resulting equation. x1 x 2 3 2 . a 1 2 6 1 b 5 x1 x 2 3 2 . c d x 1 x23 x x 2 32 2x 1 x 2 3 5 6 3x 5 9 x53 Checking this potential solution by substitution in the original equation shows that 3 makes two denominators 0. Thus, 3 cannot be a solution, so there is no solution for this equation. 4 EXERCISES 28. 3x x 27562 3 4 29. 4 8 3 2 1 50 x23 2x 1 5 x23 6.

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