College Algebra Tutorial 24


College Algebra
Answer/Discussion to Practice Problems  
Tutorial 24: Practice Test on Tutorials 14 - 23


WTAMU > Virtual Math Lab > College Algebra > Tutorial 24: Practice Test on Tutorials 14 - 23

 

 

Problems 1a - 1c:  Solve the given equation and determine whether the equation is an identity, a conditional equation or an inconsistent equation.

 
1a. problem 1a
 
check markAnswer:
ad1a

Since the variable dropped out AND it is a true statement, the answer is all real numbers, which means this equation is an identity.

 


 

1b. problem 1b
 
check markAnswer:
ad1b

Since we came up with the solution x = 29/36, this would be an example of a conditional equation.

 


 

1c. problem 1c
 
check markAnswer:
ad1c

Since the variable dropped out AND it is a false statement, the answer is no solution which means this is an inconsistent equation.

 


 

Problems 2a - 2b:  Solve the given equation.

 
2a. problem 2a
 
check markAnswer:
ad2a

Note that -2 does not cause any denominators to be zero.  So it is not an extraneous solution. 

-2 is the solution to our equation.

 


 

2b. problem 2b
 
check markAnswer:
ad2b

Note that 3 does cause two of the denominators to be zero. 

So 3 is an extraneous solution.  That means there is no solution. 

The answer is NO solution.

 


 

Problems 3a - 3c:  Solve the given word problem.

 
3a. In last week’s football game, Ralph scored 6 less than twice what Tommy scored.  The sum of their scores is 30.  How many points did Ralph and Tommy score individually?
 
check markAnswer:
Assign variable:
 x = Tommy’s score
2x - 6 = Ralph’s score

Since their sum is 30, we get the following equation:
ad3a1

Solving the equation we get:
ad3a2
 

Answer:
Tommy scored 12 points.
Ralph scored 2(12) - 6 = 18 points.

 


 

3b.  The ages of three sisters are three even consecutive integers.  If the sum of the 1st, four times the 2nd, and twice the 3rd is 86, what are the three ages?
 
check markAnswer:
Assign variable:
x = 1st even consecutive integer
x + 2 = 2nd even consecutive integer
x + 4 = 3rd even consecutive integer

Since the sum of the 1st, four times the 2nd, and twice the 3rd is 86, we get the following equation:
ad3b1

Solving the equation we get:
ad3b2
 

Answer:
The three ages are 10, 12, and 14.

 


 

3c.  You are buying a computer at a markdown price of $960.  If the markdown price was 20% off of the original price, how much was the computer originally?
 
check markAnswer:
Assign variable:
x = original price

Since the markdown price is $960, we get the following equation:
ad3c1

Solving the equation we get:
ad3c2
 

Answer:
The original price of the computer is $1200.

 


 

Problem 4a:  Solve for the given variable.

 
4a. problem 4a for y.
 
check markAnswer:
ad4a

 


 

Problems 5a - 5b: Solve the given equation by factoring.

 
5a. problem 5a
 
check markAnswer:
ad5a

There are two solutions to this equation: x = -2 and x = 7.

 


 

5b. problem 5b
 
check markAnswer:
ad5b

There are two solutions to this equation: x = -1/2 and x = -2/3.

 


 

Problems 6a - 6b:  Solve the given equation by completing the square.

 
6a. problem 6a
 
check markAnswer:
ad6a

There are two solutions to this equation: x = -1 + 2i and x = -1 - 2i.

 


 

6b. problem 6b
 
check markAnswer:
ad6b
 

There are two solutions to this equation: x = 7/4 + 1/4 = 2 and x = 7/4 - 1/4 = 6/4 = 3/2 .

 


 

Problems 7a - 7b:  Solve the given equation by using the quadratic equation.

 
7a.problem 7a
 
check markAnswer:
ad7a

 


 

7b. problem 7b
 
check markAnswer:

Write in standard form:
ad7b1

Put in quadratic formula:
ad7b2

 


 

Problems 8a - 8b:  Solve the given equation by factoring.

 
8a. problem 8a
 
check markAnswer:
ad8a
 

There are three solutions to this equation: a = 0, a = -3, and a = 3.

 


 

8b. problem 8b
 
check markAnswer:
ad8b

There are three solutions to this equation: x = 5, x = -2, and x = 2.

 


 

Problems 9a - 9b:  Solve the given equation.

 
9a. problem 9a
 
check markAnswer:
ad9a1

Checking for extraneous solutions:
ad9a2

Since we got a false statement, y = 3 is an extraneous solution.

This means there is no solution to this equation.

 


 

9b. problem 9b
 
check markAnswer:
ad9b

Both x = -1 and x = 3 check.

There are two solutions to this equation: x = -1 and x = 3.

 


 

Problem 10a:  Solve the given equation.

 
10a. problem 10a
 
check markAnswer:
ad10a

a = 8 does check.

There is one solution to this equation: a = 8.

 


 

Problems 11a - 11b:  Solve the given equation.

 
11a. problem 11a
 
check markAnswer:

Writing the equation in standard form:
ad11a1

Substitution:
ad11a2

Substitute in t and solve:
ad11a3
 

Substitute value in for t and solve for x:
ad11a4

There are two solutions to this equation: x = -343 and x = 1.

 


 

11b. problem 11b
 
check markAnswer:

Substitution:
ad11b1

Substitute in t and solve:
ad11b2
 

Substitute value in for t and solve for y:
ad11b3

There are two solutions to this equation: y = 3 and y = -1.

 


 

Problems 12a - 12b:  Solve the given equation.

 
12a. problem 12a
 
check markAnswer:

First solution:
ad12a1

Second solution:
ad12a2

There are two solutions to this equation: x = 8/3 and x = -16/9.

 


 

12b. problem 12b
 
check markAnswer:

Since the absolute value is set equal to a negative number, there is no solution.

 


 

Problems 13a - 13b:  Solve, write your answer in interval notation and graph the solution set.

 
13a. problem 13a
 
check markAnswer:
ad13a1
 

Interval notation: ad13a2

Graph: 
ad13a3

 


 

13b. problem 13b
 
check markAnswer:
ad13b1
 

Interval notation: ad13b2

Graph: 
ad13b3

 


 

Problems 14a - 14d:  Solve, write your answer in interval notation and graph the solution set.

 
14a. problem 14a
 
check markAnswer:

Use the definition of absolute value to set up:
ad14a1
 

Interval notation: ad14a2

Graph: 
ad14a3

 


 

14b. problem 14b
 
check markAnswer:

Use the definition of absolute value to set up:
ad14b1
 

Interval notation: ad14b2

Graph: 
ad14b3

 


 

14c. problem 14c
 
check markAnswer:

Isolate the absolute value:
ad14c

Since the absolute value is ALWAYS positive and in this problem it is set greater than or equal to a negative number, the answer is all real numbers.

 


 

14d. problem 14d
 
check markAnswer:

Since the absolute value is ALWAYS positive and in this problem it is set less than a negative number, the answer is no solution.

 


 

Problems 15a - 15b:  Solve, write your answer in interval notation and graph the solution set.

 
15a. problem 15a
 
check markAnswer:

Write in standard form:
ad15a1

Solve quadratic equation:
ad15a2
 

Mark off boundary points on number line:
ad15a3
Note that the two boundary points create three sections on the graph: ad15a4ad15a5
ad15a6.

Chose -1 in first interval to check:
ad15a7

Since 4 is positive and we are looking for values that cause our quadratic expression to be less than or equal to  0 (negative or 0), ad15a4would not be part of the solution.
 

Chose 0 in second interval to check:
ad15a8

Since -3 is negative and we are looking for values that cause our expression to be less than or equal to 0 (negative or 0), ad15a5 would be part of the solution.
 

Chose 4 in third interval to check:
ad15a9

Since 9 is positive and we are looking for values that cause our quadratic expression to be less than or equal to  0 (negative or 0), ad15a6would not be part of the solution.
 

Interval notation: ad15a5

Graph: 
ad15a10

 


 

15b. problem 15b
 
check markAnswer:

Set numerator equal to 0 and solve:
ad15b1

Set denominator equal to 0 and solve:
ad15b2
 

Mark off boundary points on number line:
ad15b3
Note that the two boundary points create three sections on the graph: ad15b4ad15b5, and ad15b6.

Chose -4 in first interval to check:
ad15b7

Since 1/8 is positive and we are looking for values that cause our quadratic expression to be greater than  0 (positive), ad15b4would  be part of the solution.
 

Chose 0 in second interval to check:
ad15b8

Since -3/4 is negative and we are looking for values that cause our quadratic expression to be greater than  0 (positive), ad15b5would not be part of the solution.

Chose 5 in third interval to check:
ad15b9

Since 8 is positive and we are looking for values that cause our quadratic expression to be greater than  0 (positive), ad15b6would  be part of the solution.
 

Interval notation: ad15b10

Graph: 
ad15b11


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WTAMU > Virtual Math Lab > College Algebra > Tutorial 24: Practice Test on Tutorials 14 - 23


Last revised on Jan. 15, 2010 by Kim Seward.
All contents copyright (C) 2002 - 2010, WTAMU and Kim Seward. All rights reserved.