Math 152B Practice Exam II

Please do all of the following problems.  All work and all answers must be done on your own paper.  Credit earned will be based on the steps that you show that lead to the final solution.  Good Luck!

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Problem 1:  Find the LCD for the given rational expressions and convert each rational expression into an equivalent rational expression with the LCD as the denominator.

A.     10u                7
                       ,                                        
         39v              26u2v

Solution

First find the LCD.  The two denominators are 39v and 26u2v.  Notice that

    39  =  (13)(3)    and    26  =  (13)(2)

so for the number part of the LCD we get

    (13)(3)(2)  =  78

For the variables, we take the highest power of each:  u2v, so the LCD is

        LCD  =  78u2v

Now we build the two rational expressions so that they have the 78u2v as the denominator:

         10u         2u2              7             3
                                   ,                                                        
         39v      2u2             26u2v         3

This gives

          20u3                21
                          ,                                        
        78u2v             78u2v

 

 

 

B.        3x             5 + x
                        ,                                      
         x2 - 5x         5 - x

Solution

First factor the denominators.  For the first, we factor out the GCF and for the second, we factor out the "-" so that the x-term is positive.  The denominators become

        x(x - 5) and -(x - 5)

The LCD is

        LCD  =  x(x - 5)

Now we build the fractions so that they have the LCD as their denominators:

            3x              5 + x        -x
                         ,                                                   
         x2 - 5x         5 - x         -x

This gives

             3x             -x2 - 5x
                        ,                                           
         x2 - 5x          x2 - 5x

 

  

 

Problem 2:  Perform the indicated operations and express your answer in simplest form.

A       2x2 - x - 1     x2 - 6x + 8
                           .                                            
         x2 - 5x + 4    2x2 - 3x -2

              (2x + 1)(x - 1)     (x - 4)(x - 2)
   =                                .                               =   1                   
             (x - 4)(x - 1)        (2x + 1)(x - 2)

B       x2 + 2 xy + y2      y + x
                                                                 
          x2 - x - 2              x - 2

                  x2 + 2xy + y2      x - 2
        =                              .                                    
                   x2 - x - 2             y + x

                  (x + y)(x + y)         x - 2
        =                               .                                    
                     (x - 2)(x + 1)       x + y

                     (x + y)    
        =                                               
                      (x + 1)  

C.       x2 - x - 5                   1 - 2x
                                +                                 
         x2 +3x + 2            (x + 1)(x + 2)

Solution first notice that the first denominator factors as

        x2 + 3x + 2  =  (x + 1)(x + 2)

so that the denominators are the same.  We can add the rational expressions by adding the numerators and leaving the denominators the same.  We get

            (x2 - x - 5) + (1 - 2x)
                                                    
               (x + 1)(x + 2)

Now combine like terms to get

            x2 - 3x - 4
                                      
         (x + 1)(x + 2)

 

D.            9                      6
                              -                       
         q2 - q - 2             q2 - 1

First factor the denominators to get

        q2 - q - 2  =  (q - 2)(q + 1)     and     q2 - 1  =  (q - 1)(q + 1)

Now build the fractions to get

                   9(q - 1)                                   6(q - 2)
      =                                           -                                          
           (q - 2)(q - 1)(q + 1)            (q - 2)(q - 1)(q + 1)

 

Now multiply through to get

                   9q - 9                                     6q - 12
      =                                           -                                          
           (q - 2)(q - 1)(q + 1)            (q - 2)(q - 1)(q + 1)

now subtract the numerators

            (9q - 9)  - (6q - 12)           
      =                                         
            (q - 2)(q - 1)(q + 1)   

Remember to distribute the "-" sign through both terms on the right.

            9q - 9  - 6q + 12           
      =                                         
            (q - 2)(q - 1)(q + 1)   

Combine like terms

                     3q - 3           
      =                                         
            (q - 2)(q - 1)(q + 1)   

Factor the numerator

                     3(q - 1)           
      =                                         
            (q - 2)(q - 1)(q + 1)   

and cancel the "q - 1"s

                     3        
      =                                
            (q - 2)(q + 1)  

Problem 3:  Solve the following equations

A.   x3 - 2x2 + x = 0

Solution

We factor the left hand side.  First pull out the GCF:  x

        x(x2 - 2x + 1)  =  0

Now notice that the second factor is a perfect square trinomial (square of a difference) with a = x and b = 1.  We get

        x(x - 1)2  =  0

Now use the zero factor property:

        x  =  0      or      x - 1  =  0

so

        x  =  0     or     x  =  1

B.    4x2 = 15 - 4x

Solution

First set the equation equal to 0 by adding 4x - 15 to both sides:

        4x2 + 4x - 15  =  0

Now use the AC method

        AC  =  -60

Notice that

        (10)(-6)  =  60     and     10 + (-6)  =  60

so we write

        4x2 - 6x + 10x - 15  =  0

Now factor by grouping

        2x(2x - 3) + 5(2x - 3)  =  0

        (2x + 5)(2x - 3)  =  0

Now use the zero factor property

        2x + 5  =  0    or    2x - 3  =  0

subtract 5 from both sides on the first and add 3 to both sides on the second

        2x  =  -5    or    2x  =  3

Now divide both sides by 2 for both equations

        x  =  -5/2    or    x  =  3/2

 

Problem 4:  The pressure p in pounds per square foot of a wind is directly proportional to the square of the velocity v of he wind.  If a 10-mi/hr wind produces a pressure of 0.3 lb/ft2, what pressure will a 100-mi/hr wind produce?  

Solution

Since this is a "directly proportional" problem, we have

        p  =  kv2

where k is some constant.

The first phrase of the second sentence gives us

        v  =  10    p  =  0.3

Plugging these in gives

        0.3  =  (k)(10)2  =  100k

Divide both sides by 100 to get

        k  =  0.3/100  =  0.003

Putting this back into the equation gives

        p  =  0.003v

Now the question is to find p when w  =  100.  We plug in to get

        p  =  (0.003)(100)2  =  30

So the pressure will be 30 lbs/ft2 when the wind is 100 miles per hour.

 

 

Problem 5  Solve each inequality.

A.  | 2x - 4 | + 3 > 7

Solution

First subtract 3 from both sides

        |2x - 4| > 4

Now turn this into an "or" statement

        2x - 4 = 4 or 2x - 4 = -4

        2x = 8  or  2x = 0

        x = 4  or  x = 0

Now test the point x = 1 to see if we yield a true statement

        |2(1) - 4| + 3  =  |-2| + 3  =  2 + 3  = 5

is not greater than 7.  Hence the solution is the outside two intervals:

        (- , 0) U (4, )

 

B.  | 3x - 8 | + 5  < 4

Solution

First subtract 5 from both sides of the inequality

        |3x - 8| < -1

Now notice that the absolute value can never be negative, in particular it is never less than -1.  Hence there is no solution.

 

C.  3| x - 5| <  9

Solution

First divide both sides by 3 to get 

       |x - 5|  <  3

Now turn this into an "or" equation.

        x - 5 = 3  or  x - 5 = -3

        x = 8  or  x = 2

Now test a between point such as 5:

        |(5) - 5| = 0 which is less than 3, hence the between interval is the solution.  The solution is

        (2,8)

 

Problem 6  Solve the equation

        x - 3       x - 6
                 -               =  0
       x + 1       x + 5

Solution

        Multiple all three terms by the common denominator (x + 1)(x + 5)

        (x - 3)(x + 1)(x + 5)            (x - 6)(x + 1)(x + 5)
                                           -                                    =  0
                 x + 1                                 x + 5

        (x - 3)(x + 5)  -  (x - 6)(x + 1)  =  0

        [x2 - 3x + 5x - 15]  -  [x2 - 6x + x - 6]  =  0

        [x2 + 2x - 15]  -  [x2 - 5x  - 6]  =  0

        x2 + 2x - 15  -  x2 + 5x  + 6  =  0

        7x - 9  =  0

        7x  =  9

        x  =  9/7

 

Problem 7  Solve the following for m

                             r
        S  =  1  +            
                         m

Solution

first subtract 1 from both sides

                           r
        S - 1  =             
                       m

Now multiply both sides by m

        m(S - 1)  =  r

Now divide both sides by S - 1

                          r
        m  =                    
                    S - 1

       

Problem 8  Steve can paint his house in 10 hours working by himself.  Working together, Anne and Steve can paint the house in just 6 hours.  How long would it take Anne to paint the house by herself?

Solution

Since Steve can paint the house in 10 hours, he paints the house at the rate of 1/10 houses per hour.  If Anne can paint the house in x hours, she paints at a rate of 1/x houses per hour.  The sum of the individual rates equal the total rate (1/6 houses per hour).  

              1            1         1 
                  +          =             
          10           x          6  

Now multiply by the common denominator 30x

              1 (30x)           1(30x)        1 (30x)
                          +                 =                  
             10                   x                6  

        3x + 30  =  5x

        2x  =  30

        x  =  15

It would take Anne 15 hours to paint the house by himself.

 

Problem 9  Derek bicycled 36 miles to get to Echo Summit and back and Nick bicycled 60 miles to get to Carson Pass and back.  Nick rode 3 miles per hour faster than Derek, and his trip took an hour longer than Derek's.  What is the fastest speed that Derek could have been traveling?  (You must set up the equations, that is, no guessing).

Solution

We construct the following Distance-Rate-Time table

  Distance Rate Time
Derek 36 r t
Nick 60 r + 3 t + 1

This gives us the two equations

        36  =  rt        60  =  (r + 3)(t + 1)

The first equation gives us

        t  =  36/r

so that

        60  =  (r + 3)(36/r + 1)

Multiplying both sides by r gives

        60r  =  (r + 3)(36 + r)  =  r2 + 39r + 108

or

        r2 - 21r  + 108  =  0

        (r - 9)(r - 12)  =  0

So that

        r  =  9     or      r  =  12

The fastest that Derek could have been traveling is 12 miles per hour.

 

Problem 10  Simplify

                          x
           3 -                          
                        x + 1
                                          
             x            1
                    +                      
             6            4

Solution

We first find the least common denominator.  The denominators are  x + 1, 6, and 4.  The LCD is

        LCD  =  12(x + 1)

Now multiply the two terms in the numerator and the two in the denominator by the LCD

                                      x        12(x + 1)
           3(12)(x + 1) -                                
                                   x + 1           1   
                                                                              
             x       12(x + 1)             1      12(x + 1)
                                         +                                    
             6             1                  4             1

Now cancel to get

                       3(12)(x + 1)  - 12x
            =                                             
                       2x(x + 1) + 3(x + 1)

Distribute to get

                       36x + 36  - 12x
            =                                     
                     2x2 + 2x + 3x + 3

Combine like terms

                      24x + 36
            =                           
                    2x2 + 5x + 3

Factor the numerator and the denominator

                        12(2x + 3)
            =                               
                     (2x + 3)(x + 1)

Now cancel the x + 1 to get

                      12
            =                 
                    x + 1

Problem 11  The south facing part of a small roof is to be constructed so that its length is four feet more than the vertical distance from the ceiling to its top.  Find the length of the roof if the horizontal length of the roof is 8 feet.

Pic of House with right triangle roof dimesions x, x+4, 8

Solution

Use the Pythagorean Theorem.  We have

x2 + 82  =  (x + 4)2

Now FOIL out the left hand side to get

x2 + 64  =  x2 + 4x + 4x + 16

x2 + 64  =  x2 + 8x + 16

Now subtract x2 + 64 from both sides to get

8x - 48  =  0

8x  =  48

x  =  6

We can now use the picture to answer the question.  We can conclude that the length of the small roof is 10 feet.  (6 + 4  =  10)

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