2a. 
 
Answer:

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. 
 
Answer:

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.

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?
 
Answer:
Assign variable:
 x = Tommy’s score
2x - 6 = Ralph’s score

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

Solving the equation we get:

 

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?
 
Answer:
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:

Solving the equation we get:

 

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?
 
Answer:
Assign variable:
x = original price

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

Solving the equation we get:

 

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

5a. 
 
Answer:

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

5b. 
 
Answer:

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

6a. 
 
Answer:

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

6b. 
 
Answer:

 

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

7b. 
 
Answer:

Write in standard form:

Put in quadratic formula:

8a. 
 
Answer:

 

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

8b. 
 
Answer:

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

9a. 
 
Answer:

Checking for extraneous solutions:

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

This means there is no solution to this equation.

9b. 
 
Answer:

Both x = -1 and x = 3 check.

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

10a. 
 
Answer:

a = 8 does check.

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

11a. 
 
Answer:

Writing the equation in standard form:

Substitution:

Substitute in t and solve:

 

Substitute value in for t and solve for x:

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

11b. 
 
Answer:

Substitution:

Substitute in t and solve:

 

Substitute value in for t and solve for y:

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

12a. 
 
Answer:

First solution:

Second solution:

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

12b. 
 
Answer:

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

13a. 
 
Answer:

 

Interval notation: 

Graph: 

13b. 
 
Answer:

 

Interval notation: 

Graph: 

14a. 
 
Answer:

Use the definition of absolute value to set up:

 

Interval notation: 

Graph: 

14b. 
 
Answer:

Use the definition of absolute value to set up:

 

Interval notation: 

Graph: 

14c. 
 
Answer:

Isolate the absolute value:

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. 
 
Answer:

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

15a. 
 
Answer:

Write in standard form:

Solve quadratic equation:

 

Mark off boundary points on number line:

Note that the two boundary points create three sections on the graph: , 
.

Chose -1 in first interval to check:

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), would not be part of the solution.
 

Chose 0 in second interval to check:

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),  would be part of the solution.
 

Chose 4 in third interval to check:

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), would not be part of the solution.
 

Interval notation: 

Graph: 

15b. 
 
Answer:

Set numerator equal to 0 and solve:

Set denominator equal to 0 and solve:

 

Mark off boundary points on number line:

Note that the two boundary points create three sections on the graph: , , and .

Chose -4 in first interval to check:

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

Chose 0 in second interval to check:

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

Chose 5 in third interval to check:

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

Interval notation: 

Graph: