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Virtual Math Lab

Intermediate Algebra
Answer/Discussion to Practice Problems  
Tutorial 36: Practice Test on Tutorials 32 - 35




Problem 1a: 

 
problem 1a
 

checkAnswer:

Our restriction here is that the denominator of a fraction can never be equal to 0.  So, to find our domain, we want to set the denominator "not equal" to 0 to restrict those values.
 

ad1a

Our domain is all real numbers except -9 and 9, because they both make the denominator equal to 0, which would not give us a real number answer for our function. 


 
 
Problem 2a: 

 
problem 2a
 

checkAnswer:

ad2a


 
 
 
Problem 2b: 

 
problem 2b
 

checkAnswer:
 

ad2b


 
 
Problem 2c: 

 
problem 2c
 

checkAnswer:

Step 1: Find the LCD if needed.

The first denominator has the following factor: ad2c

The second denominator, 3 - x, looks like the first denominator except the  signs are switched.  We can rewrite this as ad2c

Putting all the different factors together and using the highest exponent, we get the following LCD: ad2c3
 

Step 2: Write equivalent fractions using the LCD if needed.

Now the two fractions have a common denominator, so we do not have to  rewrite the rational expressions.
 

Step 3: Combine the rational expressions

AND 

Step 4: Reduce to lowest terms.

ad2c4


 
Problem 2d: 

 
problem 2d
 

checkAnswer:

Step 1: Find the LCD if needed.

The first denominator has the following two factors:

ad2d1

The second denominator has the following factor: ad2d2

The third denominator has the following factor: ad2d3

Putting all the different factors together and using the highest exponent, we get the following LCD: ad2d4
 

Step 2: Write equivalent fractions using the LCD if needed.

Since the first rational expression already has the LCD, then we do not need to change this fraction.

ad2d5
 

Rewriting the second expression with the LCD:

ad2d6
 

Rewriting the third expression with the LCD:

ad2d7
 

Step 3: Combine the rational expressions

AND 

Step 4: Reduce to lowest terms.

ad2d8


 
Problem 3a: 

 
problem 3a
 

checkAnswer:

Note that I used method I described in Tutorial 34 (Complex Fractions) to work this problem.  It is perfectly find to use method II here.

Step 1:   If needed, rewrite the numerator and denominator so  that they are each a single fraction

AND

Step 2:  Divide the numerator by the denominator by multiplying  the numerator by the reciprocal of the denominator

AND

Step 3: If needed, simplify the rational expression.

ad3a


 
 
 
Problem 3b: 

 
problem 3b
 

checkAnswer:

Note that I used method II described in Tutorial 34: Complex Fractions to work this problem.  It is perfectly fine to use method I here.

Rewriting it with positive exponents we get:
ad3b1

Step 1: Multiply the numerator and denominator of the overall complex fractions by the LCD of the smaller fractions

AND

Step 2: If needed, simplify the rational expression.
 

The two denominators of the numerator's fractions have the  following factors:    a and b
 

The two denominators of the denominator's fractions  have the  following factors: ad3b2 and ad3b3
 

Putting all the different factors together and using the highest exponent we get the following LCD for all the small fractions: ad3b4
 

Multiplying numerator and denominator by the LCD and simplifying we get:

ad3b5
 


 
 
Problem 4a: 

 
problem 4a
 

checkAnswer:

ad4a


 
 
Problem 4b: 

 
problem 4b
 

checkAnswer:

Step 1: Set up the long division 

AND

Step 2: Divide 1st term of divisor by first term of dividend to get first term of the quotient

AND

Step 3:  Take the term found in step 1 and multiply it times the divisor

AND

Step 4:  Subtract this from the line above

AND

Step 5:  Repeat until done

AND

Step 6: Write out the answer.

ad4b1

Final answer: 

ad4b2


 

buffalo top

 


Last revised on July 17, 2011 by Kim Seward.
All contents copyright (C) 2001 - 2011, WTAMU and Kim Seward. All rights reserved.