University:
High School
Course:
High School Mathematics
Academic year:

2022
Views:

492

Pages:

6
Author:

efreet4ifh37v

Solving Rational Equations For use with Exploration 7.5 Essential Question How can you solve a rational equation? 1 EXPLORATION: Solving Rational Equations *Hint: Exclusion asymptotes Work with a partner. Match each equation with the graph of its related system of equations. Then use the graph to solve the equation. [\frac{x+1}{x-2}=1] 1/x 1/x F [\frac{2}{x-2}=x+2] 1/x y=x+2 D [\frac{2}{x-3}=x-1] 3/x y=x-1 A [\frac{2}{x-1}=x] 1/x y=x C [\frac{1}{x-1}=-1] 1/x y=-1 B [\frac{1}{x}=x^{2}] 1/x parabola E Certainly, here's the breakdown of the image you provided: Textual Content: Notetaking with Vocabulary For use after Lesson 7.5 Extra Practice In Exercises 1-4, solve the equation by cross multiplying. Check your solution(s). [\frac{x-8}{2x-7}=\frac{5}{7}] 7(x-8) = 5(2x-7) 7x - 56 = 10x - 35 -3x = 21 x = -7 Check: [\frac{-7-8}{2(-7)-7}=\frac{5}{7}] [\frac{-15}{-21}=\frac{5}{7}] [\frac{5}{7}=\frac{5}{7}] [\frac{x+1}{x-2}=1] x + 1 = x - 2 1 = -2 No solution [\frac{x-1}{x+2}=\frac{6}{x-2}] (x - 1)(x - 2) = 6(x + 2) x^2 - 3x + 2 = 6x + 12 x^2 - 9x - 10 = 0 (x - 10)(x + 1) = 0 x = 10 or x = -1 Check: [\frac{10-1}{10+2}=\frac{6}{10-2}] [\frac{9}{12}=\frac{6}{8}] [\frac{3}{4}=\frac{3}{4}] [\frac{-1-1}{-1+2}=\frac{6}{-1-2}] [\frac{-2}{1}=\frac{6}{-3}] [-2 = -2] [\frac{x}{x+3}=\frac{x+21}{x+3}] x(x + 3) = (x + 21)(x + 3) x^2 + 3x = x^2 + 24x + 63 -21x = 63 x = -3 Check: [\frac{-3}{-3+3}=\frac{-3+21}{-3+3}] The equation is undefined for x = -3. Therefore, there is no solution.