precalc chapter 7 practice test qanda - mathguy.us · precalculus chapter 7 practice test...

PreCalculusChapter7PracticeTest Name:____________________________

Page | 1

Note the exponents of 2 in the equations. Any exponent greater than 1 will identify a system as

non‐linear. This system is non‐linear. Answer B.

Typically, you would begin solving a set of three equations by selecting pairs of equations and

eliminating the same variable in each pair. I’ll start this one by eliminating the variable .

Add 1st and 3rd equations Add 2nd and 3rd equations

1 9 3 3

2 4 2 2 12

We got lucky here because we were able to eliminate two variables at the same time by adding

the first and third equations. Normally, this would not happen, and you would have to solve the

set of two simultaneous equations which result. Let’s continue.

Solve for : Then, solve for : Then, solve for :

2 4 2 2 12 1

2 2 ∙ 2 2 12 2 4 1

4 2 12 2 1

2 8 3

4

Finally, test your results in one of the original equations, but not the one used to solve for .

Second equation: 2 4 3 9

Solution: , ,

Note: You can download a Microsoft Excel file from the following link

that will allow you to explore systems of 3 equations. Give it a try. http://www.mathguy.us/MathTidbits.php


Page | 2

For problems 3 to 5, we need only write the form of the decomposition. We do not need to solve the

decomposition for the constants , and .

This rational function has no repeated linear factors in the denominator, so the decomposition is

straightforward:

2 36 6

This rational function has a repeated linear factor 7 , so the decomposition must include

each integral exponent of 7 , up to the exponent of the term in the rational function 2 :

3 15 7

This rational function has a quadratic function in the denominator so the decomposition must

take this into account:

2 52 4


Page | 3

Write the form of the decomposition:

Multiply both sides by 1 5 : 15 39 5 1

Simplify: 15 39 5

Write the simultaneous equations and solve them:

15 5 39

Solve for : Then, solve for :

15 15

5 39 6 15

4 24

So, the partial fraction decomposition is:

15 391 5


Page | 4


Multiply both sides by 3 : 7 36 3 3

Expand and simplify: 7 36 6 9 3

7 36 6 3 9


0 6 3 7 9 36


9 36 0 6 3 7

4 0 6 4 3 4 7

24 12 7


7 363


Page | 5


Multiply both sides by 1 1 : 10 2 1 1

Expand and simplify: 10 2

10 2


0 10 2

Now, let’s eliminate form the 2nd equation with some help from the 3rd equation.

10

2

2 12

Then,


2 12 0 2

4 0 4 2 0

3 12


10 21 1


Page | 6


Multiply both sides by 4 : 3 1 4

Expand and simplify: 3 1 4 4

3 1 4 4


0 1 4 3 4 1

Solve for : Then, solve for :

4 3 4 1

4 0 3 4 1 1

4 1


3 14


Page | 7

6 8 16

8 16 6

7 10 0

5 2 0

2, 5

When 2, we get:

2 6, so 4⇒ 2, 4 is a solution

When 5, we get:

5 6, so 1⇒ 5, 1 is a solution

So, our solutions are: , , ,

15 56

15 15 56

15 56

15 56 0

7 8 0

7, 8

When 7, we get:

15 7 8⇒ 7, 8 is a solution

When 8, we get:


So, our solutions are: , , ,


Page | 8

2 66 41

Let’s use the Addition (i.e., Elimination) Method

2 66 multiply by 1 2 66

41 multiply by 1 41

25

5

When 5, we get:

5 41

25 41

16 ⇒ 4

5, 4 and 5, 4 are solutions

When 5, we get:

5 41

25 41

16 ⇒ 4


So, the entire solution set is:

, , , , , , ,

Note: you can graph multiple conic sections (and lines) using the Algebra App, available at:

http://www.mathguy.us/Apps/AboutAlgebraMainApp.php.

On the opening page, click on the “Conic Sections” button in the “More Algebra” column.

On the left hand side of the Conic Sections page, click on the “Graph Multiple Equations”

button

You may enter up to 4 equations, using either General Form or Standard Form for each.


Page | 9

29 4 17


29 multiply by 1 29

4 17 multiply by 1 4 17

4 12

4 12 0

2 6 0

2, 6

When 2, we get:

4 17

4 2 17

8 17

25 ⇒ 5


When 6, we get:

4 17

4 6 17

24 17

7

Result: no real solutions when 6

So, the entire solution set is:

, , ,


Page | 10

Let’s start by solving this for .

2 6

32

To graph this, do the following:

Graph the line: 3 .

The line will be solid because there

is an “equal sign" included in the

inequality.

Fill in the portion of the graph

above the line because of the

“greater than” portion of the

inequality.

To graph this inequality, do the following:

Graph the curve: 2 .

Some points on the curve:

2, 0.25 , 0, 1 , 2, 4

The curve will be solid because

there is an “equal sign" included in

the inequality.

Fill in the portion of the graph

below the curve because of the

“less than” portion of the

inequality.


Page | 11

is the interior of a circle with

center , and radius √ .

To graph this inequality, do the following:

Graph the circle: 49. Some points on the curve:

0, 7 , 0, 7 , 7, 0 , 7, 0

The curve will be solid because there is an

“equal sign" included in the inequality.

Fill in the interior of the circle because of

the “less than” portion of the inequality.

(orange and green areas)

Graph the line: 4. The line will be dashed because there is no


Fill in the portion of the graph below the

line because of the “less than” portion of

the inequality.

(violet and green areas)

Graph the line: 2 3. The line will be dashed because there is no


Fill in the portion of the graph above the

line because of the “greater than” portion

of the inequality.

The green area is the area of intersection of the two linear inequalities.


Page | 12


Graph the parabola: .


0, 0 , 2, 4 , 2, 4

The curve will be dashed because there is

no “equal sign" included in the inequality.

Fill in the portion of the graph above the

curve because of the “greater than”

portion of the inequality.


Put this in slope intercept form

10 6 60 6 60 10

1053

Graph the line: 10 .

The line will be solid because there is an


Fill in the portion of the graph below the line because of the “less than” portion of the

inequality.

The green area (and contiguous magenta line) is the area of intersection of the two linear

inequalities.


Page | 13


Graph the circle: 49. Some points on the curve:

0, 7 , 0, 7 , 7, 0 , 7, 0

The curve will be solid because there is

an “equal sign" included in the

inequality.

Fill in the interior of the circle because of

the “less than” portion of the inequality.


Put this in “ ” form

0

Graph the parabola: .


0, 0 , 2, 4 , 2, 4

The curve will be dashed because there is no “equal sign" included in the inequality.

Fill in the portion of the graph above the curve because of the “greater than” portion of

the inequality.

The green area is the area of intersection of the two linear (and contiguous orange curve)

inequalities.


Page | 14


5 20

2 7

3 27

3

Since the ship is in Q1, we must have positive and a

positive . So, 3. Then,

5 20

5 3 20

45 20

25 ⇒ 5 in Q1 So, , is the exact location

7 144

Let’s use the Substitution Method

7 7 144

7 144

7 144 0

16 9 0

16, 9

When 16, we get:


When 9, we get:


So, the two numbers are: and


Page | 15

37 5


5 5 37

10 25 37

2 10 12 0

5 6 0

6 1 0

6, 1

When 6, we get:


When 1, we get:


So, the two numbers are: and

2 2 42 90 with the dimensions of the rectangle being: by


21 21 90

21 90

21 90 0

15 6 0

6, 15

When 6, we get:

21 6 15⇒ 6, 15 is a solution

When 15, we get:

21 15 6⇒ 15, 6 is a solution

So, the dimensions are: ft. by ft.

precalc chapter 7 practice test qanda - mathguy.us · precalculus chapter 7 practice test...

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