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Alg 1 11.8 pg 694

Alg 1 11.8 pg 694

Alg 1 11.8 pg 694

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GUIDED PRACTICEVocabulary Check ✓Concept Check ✓Skill Check ✓1. Describe the shape of a hyperbola. What is an asymptote of a hyperbola?In Exercises 2–4, find the least common denominator.2. 1 x , x 23 1 11 1, 3 3. , , 4. , 3 x4 x 6x2 8x2x 2 +6x +9 x + 35. Which of the equations in Exercises 6–11 can you solve using cross multiplying?Explain your answer.Solve the equation. Remember to check for extraneous solutions.6. 3 x = xº 4x 2x 155 7. = 8. + = 12x + 1 x º 26 x 64 449. = 10. = 3 3x º6 x(x +1) x 11. 3 + 4 x + 4 x = º5 x 2 º2xx 2 +4xFind the center of the hyperbola. Draw the asymptotes and sketch the graph.4212. y = º 3 13. y = + 3x + 5x º 2PRACTICE AND APPLICATIONSSTUDENT HELPExtra Practiceto help you masterskills is on p. 807.CROSS MULTIPLYING Solve the equation by cross multiplying.x 714. = 5 3 15. x = 1 4 16. 4 105x = 125(x + 2)5 56 x 7 517. = 18. = 19. x = x + 4 3(x +1)x + 2 4 + 1 x º 3MULTIPLYING BY THE LCD Solve the equation by multiplying each side bythe least common denominator.20. 5 6 9º xx 9 = 21. = + 2x 2x + 9 x + 97 122. º = 2 3x º 12 x º 4 3 23. 1 1 22 + = x º 4 x + 4 x 2 º16178 x + 524. + 1 = º 25. 2 + = x º 4 x 2 + x º20x º 5 x 2 º25STUDENT HELPHOMEWORK HELPExample 1: Exs. 14–19,26–38Example 2: Exs. 20–38Example 3: Exs. 20–38Example 4: Exs. 51, 52Example 5: Exs. 39–50Example 6: Exs. 39–50Example 7: Exs. 39–50CHOOSING A METHOD Solve the equation.26. 1 4 + 4 x = 1 x 27. º 3x º 2 = 28. 1 x + 1 x º 15 º 2 1 = 5xxx 829. º 9 x = 1 9 30. x + 42 = x 31. 2 xx º x 3 = 4 8º 3 2232. = 33. + 1 x + 7 x + 2x + 3 x = 410 34. º 3 3xx + 3 5 = 1 0x+ 13x+ 935. x + 3 56 º 3x85x = 36. + 1 = x º 5 x 2 º 13x +40x + 4 x 2 º 2x º 24x222xx x37. º 1 = 38. º =2 º1 x º 11 x 2 º 5x º 66x + 3 x + 7 x 2 + 10x + 21<strong>694</strong> Chapter 11 Rational Equations and Functions

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