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solving rational equations 8 5 and inequalities teks 2a 10 d rational functions determine the solutions of rational equations using graphs tables and algebraic methods objective who uses this solve ...

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                                                               Solving Rational Equations 
                          8-5 
                                                               and Inequalities
                                TEKS  2A.10.D Rational functions: determine the solutions of rational equations 
                                using graphs, tables, and algebraic methods.
                         Objective                              Who uses this?
                         Solve rational equations               Kayakers can use rational equations to 
                         and inequalities.                      determine how fast a river is moving. 
                         Vocabulary                             (See Example 3.)
                         rational equation                     A  rational equation  is an equation 
                         extraneous solution
                         rational inequality                   that contains one or more rational 
                                                               expressions. The time t in hours that it 
                                                               takes to travel d miles can be determined 
                                                                                                    d
                                                                                                    __
                                                               by using the equation t =       , where r is the 
                                                                                                    r
                                                               average rate of speed. This equation is a 
                                                               rational equation.
                                Also 2A.2.A, 2A.10.A,          To solve a rational equation, start by 
                                2A.10.B, 2A.10.C,              multiplying each term of the equation by 
                                2A.10.E, 2A.10.F               the least common denominator (LCD) 
                                                               of all of the expressions in the equation. 
                                                               This step eliminates the denominators of 
                                                               the rational expressions and results in an 
                                                               equation you can solve by using algebra.
                                     EXAMPLE 1   Solving Rational Equations
                                                                                                      8
                                                                                                      _
                                                                     Solve the equation x +         = 6.
                                                                                      8               x
                                                                              ( )     _( )        ( )
                                                                         x x   +          x   = 6  x               Multiply each term by the LCD, x.
                                                                                      x
                                                                                      2
                                                                          x    + 8 = 6x                            Simplify. Note that x ≠ 0.
                                                                              2
                           Factoring is not the                           x     - 6x + 8 = 0                       Write in standard form.
                           only method of                                (       )(        )
                           solving the quadratic                           x - 2    x - 4    = 0 Factor.
                           equation that results                        x - 2 = 0 or x - 4 = 0                     Apply the Zero Product Property.
                           in Example 1. You 
                           could also complete                           x = 2 or x = 4                            Solve for x.
                           the square or use the 
                           Quadratic Formula.                                            8                               8
                                                                                         _                               _
                                                                     Check  x +         = 6                         x +         = 6  
                                                                                         x                               x
                                                                                   −−−−−−                          −−−−−−
                                                                                         8                               8
                                                                                         _                               _
                                                                                   2 +           6                 4 +           6
                                                                                         2                               4
                                                                       6   6 ✔ 6   6 ✔
                                                                                    Solve each equation.
                                                                                          10      4                         6     5        7                          6
                                                                                          _ _                               _ _            _                          _
                                                                1a.        =         + 2                              1b.        +          = -           1c.  x =          - 1
                                                                                           3      x                         x     4        4                          x
                                                               An  extraneous solution  is a solution of an equation derived from an original 
                                                               equation that is not a solution of the original equation. When you solve a 
                                                               rational equation, it is possible to get extraneous solutions. These values should 
                                                               be eliminated from the solution set. Always check your solutions by substituting 
                                                               them into the original equation.
                         600600    Chapter 8 Rational and Radical Functions
               aa207se_c08l05_0600_0607.indd   600207se_c08l05_0600_0607.indd   600                                                                                                     11/3/06   12:22:48 PM/3/06   12:22:48 PM
                                                                         EXAMPLE 2  Extraneous Solutions
                                                                                                                        Solve each equation.
                                                                                                                                     3x              2x + 3
                                                                                                                                  _ _
                                                                                                                         A         =                                 
                                                                                                                                   x - 3              x - 3
                                                                                                                                           3x                            2x + 3
                                                                                                                                        _( ) _( )
                                                                                                                                                      x - 3   =                            x - 3                 Multiply each term by the LCD, x - 3.
                                                                                                                                        x - 3                             x - 3
                                                                                                                                            3 x                          2x + 3
                                                                                                                                        _( ) _( )
                                                                                                                                                      x - 3   =                            x - 3    Divide out common factors.
                                                                                                                                        x - 3                             x - 3
                                                                                                                                         3x = 2x + 3                                                             Simplify. Note that x ≠ 3.
                                                                                                                                         x = 3                                                                   Solve for x.
                                                                                                                                  The solution x = 3 is extraneous because it makes the denominators of 
                                                          A rational expression                                                   the original equation equal to 0. Therefore, the equation has no solution.
                                                          is undefined for                                              Check Substitute 3 for x in the original equation.
                                                          any value of a 
                                                                                                                                                ( )              ( )
                                                          variable that makes                                                                 3  3             2  3    + 3
                                                                                                                                            __
                                                          a denominator in the                                                                               =                     
                                                                                                                                             3 - 3                3 - 3
                                                          expression equal to 0.                                                            −−−−−−−−−−−
                                                                                                                                                      9        9
                                                                                                                                                     _ _
                                                                                                                                               ✘                                                                 Division by 0 is undefined.
                                                                                                                                                      0        0
                                                                                                                                   2x - 9               x             5
                                                                                                                                  ___
                                                                                                                          B                         +           =            
                                                                                                                                    x - 7               2        x - 7
                                                                                                                                        2x - 9                                    x                                  5                                  Multiply each 
                                                                                                                                        _( ) _ ( ) _ ( ) term by the 
                                                                                                                                                         · 2  x - 7   +          · 2  x - 7   =                                · 2  x - 7   
                                                                                                                                         x - 7                                    2                              x - 7                                                    (           )
                                                                                                                                                                                                                                                              LCD, 2 x - 7  .
                                                                                                                                        2x - 9                                    x                                  5                               Divide out 
                                                                                                                                        _( ) _ ( ) _ ( )
                                                                                                                                                          · 2  x - 7   +          · 2  x - 7   =                              · 2  x - 7   
                                                                                                                                          x - 7                                   2                              x - 7                                    common factors.
                                                                                                                                                                  (              )          (            )         ( )
                                                                                                                                   2 2x - 9  + x  x - 7   = 5 2                                                                        Simplify. Note that x ≠ 7.
                                                                                                                                                                                              2                                        Use the Distributive 
                                                                                                                                   4x - 18 +  x    - 7x = 10                                                                                 Property.
                                                                                                                                                                                   2
                                                                                                                                    x    - 3x - 28 = 0                                                                                 Write in standard form.
                                                                                                                                                                             (            )(             )
                                                                                                                                    x - 7  x + 4  = 0 Factor.
                                                                                                                                                                                                                                       Use the Zero Product 
                                                                                                                                   x - 7 = 0 or x + 4 = 0 
                                                                                                                                                                                                                                             Property.
                                                                                                                                   x = 7 or x = -4  Solve for x.
                                                                                                                                  The solution x = 7 is extraneous because it makes the denominators of 
                                                                                                                                  the original equation equal to 0. The only solution is x = -4.
                                                                                                                                                   2x - 9           x           5
                                                                                                                                                   _____            __      ____
                                                                                                                        Check Write                              +          =            as  
                                                                                                                                                    x - 7           2        x - 7
                                                                                                                                       2x - 9           x          5
                                                                                                                                      _____            __       ____
                                                                                                                                                     +          -           = 0. Graph the left side of 
                                                                                                                                        x - 7           2       x - 7
                                                                                                                                      the equation as Y1 and identify the values 
                                                                                                                                      of x for which Y1 = 0.
                                                                                                                                       The graph intersects the x-axis only 
                                                                                                                                      when x = -4. Therefore, x = -4 is the 
                                                                                                                                      only solution.
                                                                                                                 
                                                                                                                                               Solve each equation.
                                                                                                                                                             16                   2                                               1                 x             x
                                                                                                                                                        __                                                                    _ _ _
                                                                                                                 2a.   2      =                                                                                      2b.                    =                 +        
                                                                                                                                                         x      - 16          x - 4                                           x - 1             x - 1             6
                                                                                                                                                                               8-5 Solving Rational Equations and Inequalities                                                     601601
                      aa207se_c08l05_0600_0607.indd   601207se_c08l05_0600_0607.indd   601                                                                                                                                                                                         11/3/06   12:22:54 PM/3/06   12:22:54 PM
                                     EXAMPLE 3  Problem-Solving Application
                                                                     A kayaker spends an afternoon paddling on a river. 
                                                                     She travels 3 mi upstream and 3 mi downstream 
                                                                     in a total of 4 h. In still water, the kayaker can 
                                                                     travel at an average speed of 2 mi/h. Based on 
                                                                     this information, what is the average speed of the 
                                                                     river’s current?
                                                                      1      Understand the Problem
                                                                     The answer will be the average speed of the current.
                                                                     List the important information:
                                                                         •  The kayaker spent 4 hours kayaking.
                                                                         •  She went 3 mi upstream and 3 mi downstream.
                                                                         •  Her average speed in still water is 2 mi/h.
                                                                       2 Make a Plan
                                                                     Let c represent the speed                                  Distance          Average              Time
                                                                     of the current. When the                                      (mi)        Speed (mi/h)              (h)
                                                                     kayaker is going upstream, 
                                                                     her speed is equal to her                                                                            3
                                                                                                                                                                       _
                                                                                                                   Up                32 - c                                     
                           distance = rate × time                    speed in still water minus c.                                                                     2 - c
                           Therefore,                                When the kayaker is going                                                                            3
                                                                                                                                                                       _
                                      distance                                                                     Down              32 + c                                     
                                     _ downstream, her speed is                                                                                                        2 + c
                           time =    rate    .
                                                                     equal to her speed in still 
                                                                     water plus c.
                                                                                       total time  =  time upstream  +  time downstream 
                                                                                                                      3                             3
                                                                                                                   _                             _
                                                                                              4         =                           +       
                                                                                                                    2 - c                         2 + c
                                                                        3 Solve
                                                                                                  3                               3                         The LCD is 
                                                                       (       )               _( )                            _( )
                                                                     4  2 - c    2 + c   =                2 - c    2 + c   +              2 - c    2 + c   
                                                                                 (       )     2 - c             (       )     2 + c             (       )      (       )        )
                                                                                                                                                                         (
                                                                                                                                                                 2 - c   2 + c  .
                                                                       (        )(        )      (       )      (       )
                                                                      4 2 - c    2 + c   = 3 2 + c   + 3 2 - c                 Simplify. Note that c ≠ ±2.
                                                                                          2
                                                                      16 - 4 c    = 6 + 3c + 6 - 3c                            Use the Distributive Property.
                                                                                          2
                                                                      16 - 4 c    = 12                                         Combine like terms.
                                                                                          2
                                                                      -4 c    = -4                                             Solve for c.
                                                                      c = ±1
                                                                     The speed of the current cannot be negative. Therefore, the average speed of 
                                                                     the current is 1 mi/h.
                                                                      4 Look Back
                                                                     If the speed of the current is 1 mi/h, the kayaker’s speed when going 
                                                                     upstream is 2 - 1 = 1 mi/h. It will take her 3 h to travel 3 mi upstream. Her 
                                                                     speed when going downstream is 2 + 1 = 3 mi/h. It will take her 1 hour to 
                                                                     travel 3 mi downstream. The total trip will take 4 h, which is the given time.
                                                                                    Use the information given above to answer the following.
                                                                                    3.  On a different river, the kayaker travels 2 mi upstream and 2 mi 
                                                                                         downstream in a total of 5 h. What is the average speed of the 
                                                                                         current of this river? Round to the nearest tenth.
                         602602    Chapter 8 Rational and Radical Functions
               aa207se_c08l05_0600_0607.indd   602207se_c08l05_0600_0607.indd   602                                                                                                      11/3/06   12:22:56 PM/3/06   12:22:56 PM
                                                                                                                                 EXAMPLE 4 Work Application
                                                                                                                                                                                                                    Jason can clean a large tank at an aquarium in about 6 hours. When Jason 
                                                                                                                                                                                                                    and Lacy work together, they can clean the tank in about 3.5 hours. About 
                                                                                                                                                                                                                    how long would it take Lacy to clean the tank if she works by herself?
                                                                                                                                                                                                                                                                            1
                                                                                                                                                                                                                                                                           _
                                                                                                                                                                                                                              Jason’s rate:         of the tank per hour
                                                                                                                                                                                                                                                                            6
                                                                                                                                                                                                                                                                          1
                                                                                                                                                                                                                                                                         _
                                                                                                                                                                                                                              Lacy’s rate:         of the tank per hour, where h is the number of hours needed 
                                                                                                                                                                                                                                                                          h
                                                                                                                                                                                                                              to clean the tank by herself
                                                                                                                                                                                                                                 Jason’s rate × hours worked    +    Lacy’s rate × hours worked    =    1 complete job   
                                                                                                                                                                                                                                                              1                                                                                                                 1
                                                                                                                                                                                                                                                             _(                    )                                                                                           _(                    )
                                                                                                                                                                                                                                       3.5                                                                                           +         3.5                                                                                               =                            1
                                                                                                                                                                                                                                                              6                                                                                                                 h
                                                                                                                                                                                                                                         1                                                  1
                                                                                                                                                                                                                                        _                      (             )             _                       (             )                 (             )
                                                                                                                                                                                                                                              (             )                                     (             )
                                                                                                                                                                                                                                          6      3.5    6h   +          3.5    6h   = 1  6h                                                                                      Multiply by the LCD, 6h.
                                                                                                                                                                                                                                                                                            h
                                                                                                                                                                                                                               3.5h + 21 = 6h Simplify.
                                                                                                                                                                                                                               21 = 2.5h  Solve for h.
                                                                                                                                                                                                                               8.4 = h
                                                                                                                                                                                                                    It will take Lacy about 8.4 hours, or 8 hours 24 minutes, to clean the tank 
                                                                                                                                                                                                                    when working by herself.
                                                                                                                                                                                                                                                        
                                                                                                                                                                                                                                                          4.  Julien can mulch a garden in 20 minutes. Together, Julien 
                                                                                                                                                                                                                                                                       and Remy can mulch the same garden in 11 minutes. 
                                                                                                                                                                                                                                                                       How long will it take Remy to mulch the garden when 
                                                                                                                                                                                                                                                                       working alone?
                                                                                                                                                                                                     A  rational inequality  is an inequality that contains one or more rational 
                                                                                                                                                                                                     expressions. One way to solve rational inequalities is by using graphs and tables.
                                                                                                                                 EXAMPLE 5   Using Graphs and Tables to Solve Rational Equations and Inequalities
                                                                                                                                                                                                                                                   x
                                                                                                                                                                                                                                          _
                                                                                                                                                                                                                    Solve                                    ≤ 2 by using a graph and a table.
                                                                                                                                                                                                                                            x - 4                                                                                                                                                                     £ä
                                                                                                                                                                                                                              Use a graph. On a graphing 
                                                                                                                                                                                                                                                                                                                       x                                                                                                                                  ­n]ÊÓ®
                                                                                                                                                                                                                                                                                                                ____
                                                                                                                                                                                                                                        calculator, let Y1 =                                                                     and 
                                                                                                                                                                                                                                                                                                                 x - 4                                                     
                                                                                                                                                                                                                                        Y2 = 2.                                                                                                                                 ££°Ó                                                                                        £™°Ó
                                                                                                                                                                                                                                        The graph of Y1 is at or below                                                                                                                                                                                6iÀ̈V>Ê>Ãޓ«ÌœÌi\
                                                                                                                                                                                                                                                                                                                                                                                                                                                                                 
                                                                                                                                                                                                                                                                                                                                                                                                                                                                           ÝÊ          {
                                                                                                                                                                                                                                                                                                                                                                                                                    
                                                                                                      The solution x < 4                                                                                                                the graph of Y2 when x < 4 or                                                                                                                                                    £ä
                                                                                                      or x ≥ 8 can be                                                                                                                   when x ≥ 8.
                                                                                                      written in set-builder 
                                                                                                      notation as                                                                                                             Use a table. The table shows that 
                                                                                                      ⎧                                                              ⎫
                                                                                                               x | x < 4 	 x ≥ 8       
                                                                                                      ⎨                                                              ⎬                                                                  Y1 is undefined when x = 4 and 
                                                                                                                                                                      
                                                                                                      ⎩                                                              ⎭                                                                  that Y1 ≤ Y2 when x < 4 or when 
                                                                                                                                                                                                                                        x ≥ 8.
                                                                                                                                                                                                                    The solution of the inequality is 
                                                                                                                                                                                                                    x < 4 or x ≥ 8.
                                                                                                                                                                                                                                                          Solve by using a graph and a table.
                                                                                                                                                                                                                                                                                   x                                                                                                                           8
                                                                                                                                                                                                                                                                           _                                                                                                                           _
                                                                                                                                                                                                      5a.         ≥ 4                                                                                                                                                                  5b.                                 = -2
                                                                                                                                                                                                                                                                            x - 3                                                                                                                       x + 1
                                                                                                                                                                                                                                                                                                                     8-5 Solving Rational Equations and Inequalities                                                                                                                                  603603
                                      aa207se_c08l05_0600_0607.indd   603207se_c08l05_0600_0607.indd   603                                                                                                                                                                                                                                                                                                                                                                                          11/3/06   12:22:59 PM/3/06   12:22:59 PM
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...Solving rational equations and inequalities teks a d functions determine the solutions of using graphs tables algebraic methods objective who uses this solve kayakers can use to how fast river is moving vocabulary see example equation an extraneous solution inequality that contains one or more expressions time t in hours it takes travel miles be determined by where r average rate speed also start b c multiplying each term e f least common denominator lcd all step eliminates denominators results you algebra x multiply simplify note factoring not write standard form only method quadratic factor apply zero product property could complete for square formula check derived from original when possible get these values should eliminated set always your substituting them into chapter radical aase cl indd se pm divide out factors...

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