1

STANDARDS OF LEARNING

CONTENT REVIEW NOTES

ALGEBRA I-Part I

4th Nine Weeks, 2018-2019

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OVERVIEW

Algebra I Content Review Notes are designed by the High School Mathematics Steering Committee as a resource

for students and parents. Each nine weeks’ Standards of Learning (SOLs) have been identified and a detailed

explanation of the specific SOL is provided. Specific notes have also been included in this document to assist

students in understanding the concepts. Sample problems allow the students to see step-by-step models for

solving various types of problems. A “ ” section has also been developed to provide students with the

opportunity to solve similar problems and check their answers. The answers to the “ ” problems are found

at the end of the document.

The document is a compilation of information found in the Virginia Department of Education (VDOE)

Curriculum Framework, Enhanced Scope and Sequence, and Released Test items. In addition to VDOE

information, Prentice Hall textbook series and resources have been used. Finally, information from various

websites is included. The websites are listed with the information as it appears in the document.

Supplemental online information can be accessed by scanning QR codes throughout the document. These will

take students to video tutorials and online resources. In addition, a self-assessment is available at the end of the

document to allow students to check their readiness for the nine-weeks test.

The Algebra I Blueprint Summary Table is listed below as a snapshot of the reporting categories, the number of

questions per reporting category, and the corresponding SOLs.

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4

Functions A.7 The student will investigate and analyze function (linear and quadratic) families and

their characteristics both algebraically and graphically, including a) determining whether a relation is a function;

The set of all solutions of an equation forms the graph of an equation. You can determine if a relation is a function if every element in the domain is paired with exactly one element in the range.

Relation Function Not a Function

T-Chart or Table x y

1 −2

3 4

5 10

x y

2 −1

2 −3

2 −5 Graph

Passes the Vertical Line Test

Does Not Pass the Vertical Line Test

Mapping

Real World A grocery store is selling peaches

for $0.98 per pound. Brandon buys 4 old CDs for

$8.00, and Sharon buys 3 old

CDs for $6.00. Pete talks to the

seller and makes a deal to buy

4 old CDs for $7.00.

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Relations and Functions

1. Draw a T-Chart for the relation (Set of Ordered Pairs) and determine whether it is a function. (4, 4), (−3, 4), (0, 4). 2. Which is the graph of a function and why?

A. B. 3. Draw a Mapping for the following relation. Is it a function?

x y

3 -4

-2 0

3 5

4. Jalen is buying tickets for the basketball game at $8 each. Is this a function?

Scan this QR code to go to a video tutorial on determining

whether a relation is a function.

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A.7 The student will investigate and analyze function (linear and quadratic) families and

their characteristics both algebraically and graphically, including b) domain and range;

T-Chart or Table

Domain Range 1 −2

3 4

5 10

Ordered Pairs

(1,-2), (3, 4), (5, 10)

Domain {1, 3, 5}

Range {-2, 4, 10}

Mapping Domain Range

Domain

Read the line Left to Right

{−𝟒, 𝟏, 𝟓}

Graph Example #1

Range Read the line

Bottom to Top

{−𝟑, 𝟐, 𝟕)} Domain

Read the line Left to Right

All real numbers

{𝑥: − ∞ < 𝑥 < ∞}

Graph Example #2

Range Read the line

Bottom to Top

All Real Numbers {𝑦: − ∞ < 𝑦 < ∞}

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Domain & Range

1. What is the domain and range for the set of ordered pairs? {(−2, 4), (0, 3), (5, −6)} 2. What is the domain and range given the graph of this function?

A. B.

Scan this QR code to go to a video tutorial on determining domain and range of a linear

function.

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Functions and Graphs A.6 The student will

a) determining the slope of a line when given an equation of the line, the graph of the line, or two points on the line.

b) writing the equation of a line when given the graph of the line, two points on the line, or the slope and a point on the line.

The set of all solutions of an equation forms the graph of an equation. As an example, if the equation is 𝑥 + 𝑦 = 5 , you can probably think of a lot of values that could satisfy x and y. Some examples are listed below in the t-chart.

x y

5 0

0 5

2 3

4 1

7 -2

To graph an equation, you can simply create a table of values by choosing values to plug in for x, and solve for the resulting y-values. Once you have x and y values, you can plot the ordered pairs and connect them with a line or curve that best fits.

Example 1: Graph 𝑦 = 2𝑥2 − 3

𝑥 𝑦 = 2𝑥2 − 3 (𝑥, 𝑦)

−2 𝑦 = 2(−2)2 − 3 = 2(4) − 3 = 5 (−2, 5)

−1 𝑦 = 2(−1)2 − 3 = 2(1) − 3 = −1 (−1, −1)

0 𝑦 = 2(0)2 − 3 = 2(0) − 3 = −3 (0, −3)

1 𝑦 = 2(1)2 − 3 = 2(1) − 3 = −1 (1, −1)

2 𝑦 = 2(2)2 − 3 = 2(4) − 3 = 5 (2, 5)

We will now plot each of these points on the coordinate plane to reveal the graph of 𝑥 + 𝑦 = 5. We will find that this graph forms a straight line, because this is a linear equation.

This line passes through all of the solutions to this linear equation, including the ones that we did not plot, such as (3, 2), (1, 4), (-1, 6) and (6, -1).

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Example 2: Graph 𝑦 − 4 = −1

2𝑥

You could start by transforming the equation for y.

𝑦 − 4 = −1

2𝑥

+4 + 4

𝑦 = −1

2𝑥 + 4

Choose x-values that will work out nicely with the fraction (i.e. even numbers in this case)

𝑥 𝑦 = −1

2𝑥 + 4 (𝑥, 𝑦)

−4 𝑦 = −1

2(−4) + 4 = 2 + 4 = 6 (−4, 6)

−2 𝑦 = −1

2(−2) + 4 = 1 + 4 = 5 (−2, 5)

0 𝑦 = −1

2(0) + 4 = 0 + 4 = 4 (0, 4)

2 𝑦 = −1

2(2) + 4 = −1 + 4 = 3 (2, 3)

4 𝑦 = −1

2(4) + 4 = −2 + 4 = 2 (4, 2)

Functions and Graphs 1. Graph 2𝑥 − 𝑦 = 3 2. Complete the function table for the function 𝑦 = |2𝑥| − 5 3. Which of these is closest to the graph of 𝑦 = −3𝑥 + 4

x y

-1

0

1

Scan this QR code to go to a video tutorial on graphing

equations using a T-Chart.

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Slope A.6 The student will

a) determining the slope of a line when given an equation of the line, the graph of the line, or two points on the line.

The slope of a line is determined by the vertical change divided by the horizontal change (or rise over run). Slope can be positive, negative, zero, or undefined.

Positive Negative Zero Undefined

𝑚 = 3

2

𝑚 = −3

2

𝑚 = 0

𝑚 𝑖𝑠 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑

You can determine slope by counting rise over run, or using the formula: 𝒎 = 𝒚𝟐−𝒚𝟏

𝒙𝟐− 𝒙𝟏

Example 1: Find the slope of the line that passes through (-3, 9) and (2, 4). 𝑥1 𝑦1 𝑥2 𝑦2

𝑚 = 4−9

2−(−3)=

−5

5= −1

Example 2: Determine the slope of the line graphed below. Example 3: Find the slope of the line that passes through (-4, 3) and (-4, 0). 𝑥1 𝑦1 𝑥2 𝑦2

𝑚 = 0−3

−4−(−4)=

−3

0= 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑

By counting rise over run, we can see that the graph goes up

5, and to the right 2.

𝑚 =5

2

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You may also be asked to find the missing coordinate of a point given the slope and a different point on the line. To do this, plug in all of the given values into the slope formula, and solve for the missing value. Example 4: The slope of the line passing through (𝑟, 8) and (5, 10) is −2. Solve for 𝑟. (𝑟, 8) and (5, 10) 𝑥1 𝑦1 𝑥2 𝑦2 −2(5 − 𝑟) = 1 ∙ 2

−10 + 𝑟 = 2 +10 + 10 𝑟 = 12

Example 5: The slope of the line passing through (6, −2) and (2, 𝑟) is 1

3. Solve for 𝑟.

(6, −2) and (2, 𝑟) 𝑥1 𝑦1 𝑥2 𝑦2 3(𝑟 + 2) = 1 ∙ (−4)

3𝑟 + 6 = −4 −6 − 6 3𝑟 = −10

÷ 3 ÷ 3

𝑟 = −10

3

𝑚 = −2

1 𝑚 =

𝑦2 − 𝑦1

𝑥2 − 𝑥1

−2

1=

10 − 8

5 − 𝑟 Now you can cross multiply!

𝑚 = 1

3 𝑚 =

𝑦2 − 𝑦1

𝑥2 − 𝑥1

1

3=

𝑟 − (−2)

2 − 6 Now you can cross multiply!

Scan this QR code to go to a

video tutorial on slope.

Write the slope in a ratio format.

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Slope 1. Find the slope of the line that passes through (4, 4) and (−3, 4). 2. Find the slope of the line that passes through (−2, 0) and (1, −5).

3. The slope of the line passing through (0, 0) and (𝑟, −5) is 5

7. Solve for 𝑟.

4. The slope of the line passing through (−2, 𝑟) and (−1, 5) is 3. Solve for 𝑟. A.7 The student will investigate and analyze function (linear and quadratic) families and

their characteristics both algebraically and graphically, including d) intercepts;

Slope-Intercept and Transformations If you are asked to translate (or shift) a line, just move two points on the line the specified units and draw a new line. Remember, the new line should be parallel to the original line. This means the slope should remain the same. Only the y-intercept will change.

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Example 1: Shift the line 𝑦 = 1

2𝑥 + 4 down 5 units and right 2 units.

Example 2: The function 𝑓(𝑥) is displayed on the graph below. Graph the function 𝑓(𝑥) + 3.

A.6 The student will

b) writing the equation of a line when given the graph of the line, two points on the line, or the slope and a point on the line.

Choose two points on the line 𝑦 =

1

2𝑥 + 4. From each point, count down

5 units and right 2 units. Plot a new point. Then, draw a line through the two new points. The equation of the

translated line is 𝑦 =1

2𝑥 − 2. Notice

that the slopes of the lines are the same.

Choose two points on the function f(x), which is 𝑦 = 2𝑥 − 2. From each

point, count up 3 units. Plot a new point. Then, draw a line through the two new points. The equation of the translated line is 𝑦 = 2𝑥 + 1. Notice that the slopes of the lines are the

same.

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Slope-Intercept

The intercept of a graph is where it crosses the axis. The x-intercept is where a graph crosses the x-axis, and the y-intercept is where a graph crosses the y-axis.

A special form of a linear equation is called slope-intercept form: 𝒚 = 𝒎𝒙 + 𝒃 . Where 𝒎 is the slope of the line and 𝒃 is the y-intercept (0, 𝑏).

Example 1: What are the slope and y-intercept of 𝑦 = 1

2 𝑥 + 4 ?

𝑦 = 𝑚𝑥 + 𝑏 The slope is the coefficient of x, and the y-intercept is the constant.

The slope of the line is 1

2 , and the y-intercept is 4.

x-intercept

y-intercept

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Example 2: What is the equation of the line with a slope of −2 and a y-intercept of −5?

𝑦 = 𝑚𝑥 + 𝑏 𝑚 = −2 𝑎𝑛𝑑 𝑏 = −5

𝑦 = −2𝑥 − 5 It is often easier to graph an equation when it is written in slope-intercept form. You can start by plotting a point on the y-axis for the y-intercept, and then count the slope as rise over run, from the y-intercept.

Example 3: Graph the equation 𝑦 = −1

3𝑥 + 2

Sometimes, you will be asked to graph an equation that is not in slope-intercept form. You can transform an equation into slope-intercept form by solving for y.

Scan this QR code to go to a video tutorial on graphing linear

equations in slope-intercept form.

To graph the equation above, put a point on the y-axis at 2, and then count the slope by going

down 1 and to the right 3. Put a second point there and draw a

line through the two points.

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Example 4: Put in slope-intercept form, state the slope and the y-intercept, and graph.

12𝑥 − 3𝑦 = 9

12𝑥 − 3𝑦 = 9

−12𝑥 − 12𝑥

−3𝑦 = −12𝑥 + 9

÷ (−3) ÷ (−3)

𝑦 = 4𝑥 − 3

The slope is 4, and the y-intercept is (0, -3).

You can find the equation of a line given two points on the line by first finding the slope. Once you have the slope and one point, you can plug this information into slope-intercept form and solve for the y-intercept (𝑏). Example 5: What is the equation of the line that passes through (0, −3) and (2, 6) ? First find the slope: (0, −3) (2, 3) 𝑥1 𝑦1 𝑥2 𝑦2

𝑚 = 𝑦2−𝑦1

𝑥2−𝑥1=

3−(−3)

2−0=

6

2=

3

1

Now you can use the slope and either of the points to solve for b.

𝑚 = 3 (0, −3) 𝑥 𝑦

𝑦 = 𝑚𝑥 + 𝑏 −3 = 3(0) + 𝑏

−3 = 0 + 𝑏 −3 = 𝑏

Now that you have 𝑚 and 𝑏, you can write an equation in slope intercept form.

𝑦 = 3𝑥 − 3

To graph the equation above, put a point on the y-axis at -3, and

then count the slope by going up 4 and to the right 1. Put a second

point there and draw a line through the two points.

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Point-slope form example: Slope (m) = 5 Point: (3,2) Step 1: Substitute the slope for m and the point for 𝒙𝟏 and 𝒚𝟏 y – 2 = 5 (x – 3) Step 2: Distribute the 5 throughout the parentheses y – 2 = 5x – 15 Step 3: Solve for y on the left hand side by adding 2 to both sides y -2 +2 = 5x – 15 + 2 Step 4: Simplify y = 5x - 13

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Slope intercept and transformations 1. What is the slope and y-intercept of 𝑦 = −7𝑥 + 12 ?

2. What is the equation of a line whose y-intercept is −4 and slope is 2

5 ?

3. What is the equation of the line that passes through (0, 1) and whose slope is 1 ?

4. What is the y-intercept of the equation 𝑥 + 3𝑦 = −9 ?

5. Graph 𝑦 = 2𝑥 − 4

6. Which of these equations has a slope of -2 ?

A. 2𝑥 − 𝑦 = 7 B. 𝑦 + 2𝑥 = 4 C. 𝑦 – 2𝑥 = 6 D. 2𝑦 − 2𝑥 = 3

7. What is the equation of the line that passes through (3, 4) and has a slope of -2 ?

8. What is the equation of the line that passes through (3, -2) and (9, -4)?

9. Graph 2𝑦 = −4𝑥 + 6

10. Graph 3𝑥 + 6𝑦 = 6

11. The function 𝑓(𝑥) is displayed on the graph below. What is the equation of the

function (𝑥) ?

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Direct and Inverse Variation A.8 The student, given a data set or practical situation, will analyze a relation to

determine whether a direct or inverse variation exists, and represent a direct variation algebraically and graphically and an inverse variation algebraically.

If the ratio between two variables is a constant, then a direct variation exists. A direct variation can be written in the form 𝑦 = 𝑘𝑥, where 𝑘 is the constant of variation. If the product of two variables is a constant, then an inverse variation exists. An

inverse variation can be written in the form 𝑦 = 𝑘

𝑥 or 𝑥𝑦 = 𝑘.

Example 1: Determine if each relation is a direct variation, inverse variation, or neither.

To write an equation of a direct variation, use a given point (other than (0, 0)) to plug into 𝑦 = 𝑘𝑥, to solve for 𝑘.

To write an equation of an inverse variation, use a given point to plug into 𝑥𝑦 =𝑘, to solve for 𝑘.

x y

1 3

2 4

3 5

x y

-1 2

0 0

1 -2

x y

3 4

1 12

-2 -6

First check the ratios:

Does the ratio 1

3=

2

4 ? NO! Therefore this is NOT a direct variation!

Next check the products: Does 1 ∙ 3 = 2 ∙ 4 ? NO! Therefore this is NOT an inverse variation!

First check the ratios:

Does the ratio 3

4=

1

12 ? NO! Therefore this is NOT a direct variation!

Next check the products: Does 3 ∙ 4 = 1 ∙ 12? Does this also equal −2 ∙ −6 ? YES! Therefore this IS an inverse variation!

First check the ratios:

Does the ratio −1

2=

1

−2 ? YES! Therefore this IS a direct variation!

Notice that we did not use the ordered pair (0,0) to check the ratios. It is impossible to divide by zero, therefore we used the other points.

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Example 2: Suppose 𝑦 varies directly with 𝑥, and 𝑦 = 17 when 𝑥 = 34. What direct variation equation relates 𝑥 and 𝑦?

Start with 𝑦 = 𝑘𝑥. We are given a value for 𝑥 and 𝑦, so plug those in and solve for 𝑘.

17 = 𝑘(34) ÷ 34 ÷ 34

1

2= 𝑘

𝑦 =1

2𝑥

Once you have a direct variation equation, you can use that equation to determine other values. Example 3: The distance that you jog, 𝑗, varies directly with the amount of time you jog, 𝑡. If you can jog 9 miles in 1.5 hours, how long will it take you to jog 4 miles? Jogging varies directly with time → 𝑗 = 𝑘𝑡

Now we need to solve for k in order to write a direct variation equation. Use the values that are related to one another. 9 = 𝑘(1.5) ÷ 1.5 ÷ 1.5 6 = 𝑘

𝑗 = 6𝑡

Now, we can use this equation to solve for the time it takes to jog 4 miles. We are given that you jog 4 miles. This will be plugged in for j. Then, solve for t. 4 = 6𝑡

÷ 6 ÷ 6 2

3 = 𝑡

Therefore, it would take you 2

3 of an hour to jog 4 miles.

This is the constant of variation!

This is the direct variation equation!

This is the constant of variation!

This is the direct variation equation!

Scan this QR code to go to a video tutorial on direct and

inverse variations.

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Direct and Inverse Variation 1. Determine if each of the following relations are a direct variation, an inverse

variation, or neither. 2. Suppose 𝑦 varies directly 𝑥, and 𝑦 = 24 when with

𝑥 = 4. What direct variation equation relates 𝑥 and 𝑦? 3. Suppose 𝑦 varies directly with 𝑥, and 𝑥 = 9 when 𝑦 = 2. What will 𝑦 be when 𝑥 =

22.5 ? 4. The amount of money spent at the mall varies directly with the amount of time you

spend shopping. If you spend $90 when you are in the mall for 2.5 hours, how much time did you spend in the mall when you spent $340?

5. Jason’s earnings (𝑒) during his summer job is directly proportional to the amount of hours he worked (ℎ). When ℎ = 4, 𝑒 = 10, what is the constant of variation?

6. The time it takes to complete a job (𝑡) is inversely proportional to the amount of workers assigned to the job (𝑤). What value would accurately represent this relationship?

x y

2 4

0 0

1 8

x y

-4 -20

5 16

8 10

Time (t) Amount of Workers (w)

3 10

5 ?

15 2

A B

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Systems of Equations A.4 The student will solve

d) systems of two linear equations in two variables algebraically and graphically; and e) practical problems involving equations and systems of equations.

A system of equations is two or more equations, whose solution is any point that each of the equations has in common. This can be seen on a graph as the intersection point of the lines. Systems of two linear equations can have no solutions, one solution, or infinitely many solutions.

No Solutions One Solution Infinitely Many Solutions

Two lines that are parallel.

These lines have the same

slope, but different y-intercepts. They will

never intersect. Therefore, there is no solution.

Two lines that intersect.

These lines have different slopes, which causes them to intersect in one place. Therefore, there is one

solution. In this example, the solution is (2, 3).

Two lines that are the same.

These lines have the same

slope and the same y-intercept. This means

they are the same line and will share all points. Therefore, there are

infinitely many solutions.

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Example 1: Systems of equations can also be solved algebraically by substitution or elimination. It is often easier to use substitution when one of the equations has a variable on the side by itself. If this is the case, you can substitute the ‘value’ of that variable into the other equation. This will allow you to solve for one variable. Example 2: Solve the system of equations by substitution.

𝑥 = 2𝑦

2𝑥 + 3𝑦 = 14

Since 𝑥 = 2𝑦, you can replace the 𝑥 in the second equation with 2𝑦!

2(2𝑦) + 3𝑦 = 14

4𝑦 + 3𝑦 = 14

7𝑦 = 14

÷ 7 ÷ 7

𝑦 = 2

Remember that the solution to a system of equations is an ordered pair! You have a y-value, so use that to help you solve for x.

𝑥 = 2(2) 𝑥 = 4

(4, 2)

What is the solution to the system of equations

pictured here?

The graphs intersect at the point (−2, −1).

Therefore, the solution is 𝑥 = −2 𝑎𝑛𝑑 𝑦 = −1.

Remember that an ordered pair is always (x, y)!

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Example 3: Solve the system of equations by substitution.

𝑦 = 3𝑥 − 6

𝑦 = 𝑥 − 10

You can substitute the ‘value’ of y from the first equation into the second equation.

3𝑥 − 6 = 𝑥 − 10

−𝑥 − 𝑥

2𝑥 − 6 = −10

+6 + 6

2𝑥 = −4

÷ 2 ÷ 2

𝑥 = −2

Remember that the solution to a system of equations is an ordered pair! You have an x-value, so use that to help you solve for y.

𝑦 = (−2) − 10 𝑦 = −12

(−2, −12)

Another method of solving a system of equations is called elimination. This is often easier when both equations are written in standard form (𝐴𝑥 + 𝐵𝑦 = 𝐶). To use the elimination method, you will add or subtract the two equations, or some multiple of them, to get one of the variables to cancel out. Example 4: Solve this system of equations by elimination.

2𝑥 + 3𝑦 = 9 4𝑥 − 3𝑦 = 15

2𝑥 + 3𝑦 = 9

4𝑥 − 3𝑦 = 15

6𝑥 = 24

÷ 6 ÷ 6

𝑥 = 4

2(4) + 3𝑦 = 9 8 + 3𝑦 = 9 −8 − 8 3𝑦 = 1 ÷ 3 ÷ 3

𝑦 = 1

3

Notice that the y terms have equal and opposite coefficients!

If you add these two equations together the y-term will cancel out! + 3𝑦 + (– 3𝑦) = 0

Don’t forget to solve for y! Plug the x-value back into one of the original equations.

(4 ,1

3 )

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You can always check your work when solving a system of equations by transforming both equations and graphing them in your calculator. The two lines should intersect at the ordered pair that you found. Below, you will see the calculator check for Example 4.

You can also check your work by plugging your values in for x and y to verify that both equations are true for those values. Sometimes the system of equations will not have variables that immediately cancel out (like the y-term did in Example 4). When this happens, you may have to multiply one or both of the equations by a constant to get two variables to have equal and opposite coefficients. Example 5: Solve this system of equations by elimination.

−2𝑥 + 15𝑦 = 10 4𝑥 + 5𝑦 = 15

2(−2𝑥 + 15𝑦 = 10) − 4𝑥 + 30𝑦 = 20 4𝑥 + 5𝑦 = 15 4𝑥 + 5𝑦 = 15

35𝑦 = 35

÷ 35 ÷ 35

𝑦 = 1

4𝑥 + 5(1) = 15 4𝑥 + 5 = 15 −5 − 5 4𝑥 = 10 ÷ 4 ÷ 4

𝑥 = 10

4=

5

2

Scan this QR code to go to a video tutorial on solving systems

of equations.

Notice that the x terms have opposite coefficient signs. What can you do to

make the coefficients equal?

Multiply the first equation by 2, then the x-terms will have equal and

opposite coefficients. +

Don’t forget to solve for x! Plug the y-value back into one of the original equations.

( 5

2 , 1)

The calculated intersection is (4,0. 3̅ )

Which is the same thing as

(4 ,1

3 )

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Example 6: Solve this system of equations by elimination.

3𝑥 + 2𝑦 = −2 4𝑥 + 3𝑦 = 1

3(3𝑥 + 2𝑦 = −2) 9𝑥 + 6𝑦 = −6 2( 4𝑥 + 3𝑦 = 1) 8𝑥 + 6𝑦 = 2

𝑥 = −8

3(−8) + 2𝑦 = −2 −24 + 2𝑦 = −2

+24 + 24 2𝑦 = 22 ÷ 2 ÷ 2 𝑦 = 11 Systems of equations are often presented as word problems. In these cases, you will often not be given the equations, and you will be responsible for setting those up. Once you have two equations set up, you can solve the system of equations using any method that you prefer.

What can you do to make two of the coefficients equal?

Multiply the first equation by 3, and the second equation by 2, then the y-terms will have equal coefficients.

_

Don’t forget to solve for y! Plug the x-value back into one of the original equations.

( −8 , 11)

Now you can subtract to get the y’s to cancel!

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Example 7: A class of 148 students went on a field trip. They took 10 vehicles, some cars and some buses. Find the number of cars and buses they took if each car holds 4 students and each bus holds 40 students.

𝑐 + 𝑏 = 10 4𝑐 + 40𝑏 = 148

−4(𝑐 + 𝑏 = 10) − 4𝑐 − 4𝑏 = −40 4𝑐 + 40𝑏 = 148 4𝑐 + 40𝑏 = 148

36𝑏 = 108

÷ 36 ÷ 36

𝑏 = 3

𝑐 + 3 = 10 −3 − 3 𝑐 = 7 Don’t forget that you can check your work by graphing! Just solve both equations for one of the variables.

We know total number of vehicles, and total number of students. We can set up two equations where those are our totals.

𝑐 + 𝑏 = 10 The number of cars plus the number of

busses equals 10 total vehicles.

4𝑐 + 40 𝑏 = 148 4 students per car plus 40 students per bus equals 148 total students.

Now we have two equations. We can solve this system using any method that we’ve learned.

Multiply the first equation by −4, then the c-terms will have equal

and opposite coefficients. +

So now we know that they took 3 busses. We can plug this value into one of the other equations to solve for the number of cars!

They took 3 busses and 7 cars!

Scan this QR code to go to a video tutorial on solving systems

of equations word problems.

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Example 8: Lauren is raising pot-bellied pigs and ostriches for fun. Among her animals, she has 17 heads and 56 legs in all. How many of each animal does she have?

𝑝 + 𝑜 = 17

4𝑝 + 2𝑜 = 56 −2(𝑝 + 𝑜 = 17) − 2𝑝 − 2𝑜 = −34 4𝑝 + 2𝑜 = 56 4𝑝 + 2𝑜 = 56

2𝑝 = 22

÷ 2 ÷ 2

𝑝 = 11

11 + 𝑜 = 17

−11 − 11 𝑜 = 6

Systems of Equations Solve each system using whatever method you prefer. 1. 3𝑥 + 𝑦 = 6 𝑦 = 𝑥 − 4

2. 3𝑥 + 4𝑦 = 11 −3𝑥 + 𝑦 = 19 3. 3𝑥 + 2𝑦 = 11 4𝑥 + 𝑦 = 18 4. The admission fee at a small fair is $1.75 for children and $3.00 for adults. On a

certain day, 1700 people enter the fair and $3375 is collected. How many children and how many adults attended?

5. Kris spent $144 on shirts. Dress shirts cost $19 and t-shirts cost $7. If he bought a total of 12, then how many of each kind did he buy?

We know total number of heads, and total number of legs. We can set up two equations where those are our totals.

𝑝 + 𝑜 = 17 Each pig has one head plus each

ostrich has one head equals 17 total heads.

4𝑝 + 2𝑜 = 56 4 legs per pig, plus 2 legs per ostrich

equals 56 legs total.

Be careful using 𝑜 as a variable! Don’t confuse it with 0. Now we have two equations. We can solve this equation using any method that we’ve learned.

Multiply the first equation by −2, then the 𝑜-terms will have equal

and opposite coefficients. +

So now we know that she has 11 pigs. We can plug this value into one of the other

equations to solve for the number of ostriches! She has 11 pigs and 6 ostriches!

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Scatterplots A.9 The student will collect and analyze data, determine the equation of the curve of best fit in

order to make predictions, and solve practical problems, using mathematical models of linear and quadratic functions.

A scatterplot is a graph made of ordered pairs relating two sets of data. Scatterplots can show trends in data.

Positive Correlation Negative Correlation No Correlation

As the x-value increases, the y-value also increases.

As the x-value increases, the y-value decreases.

The x and y-values do not appear to have any relation.

When a scatterplot has a positive or negative correlation, a trend line can be drawn to help predict other values on the line. The trend line will be the line that best fits the given data. You could estimate this trend line’s equation, called a line of fit, by selecting two points that lie on or very near the trend line, finding a slope and calculating a y-intercept.

Trend lines are shown here for a positive and negative correlation.

30

Example 1: The scatterplot below compares hours watching television and GPA. Estimate the equation of the line of fit. Then, use this equation to determine the GPA that you could expect if you watched 17 hours of TV each week.

So, to answer the question, an estimated trend line would be 𝑦 = −5𝑥 + 29 , and if you watched 17 hours of TV a week, your estimated GPA would be 2.4.

There is also a way to calculate a line of best fit using your calculator. Example 2: The cost of a gallon of gas for the past 6 years is given. Write an equation for the line of best fit, and then use this equation to predict gas prices in 2017.

Scan this QR code to go to a video tutorial on

scatterplots and line of fit.

First select two points that are either on the trend line or very near it, such as:

(2.8, 15) and (3.2, 13)

Use these points to find the slope of the line.

𝑚 = 𝑦2 − 𝑦1

𝑥2 − 𝑥1=

13 − 15

3.2 − 2.8=

−2

0.4= −5

You can then use the slope and one point to

solve for b, and write the line of fit. 𝑚 = −5 (2.8, 15)

𝑦 = 𝑚𝑥 + 𝑏

15 = (−5)(2.8) + 𝑏 𝑏 = 29

𝑦 = −5𝑥 + 29

To answer the second part of the question, we will use the equation that we just found. We are given the hours watching TV and are asked to find a GPA. “TV hours” represent the y-value, so we will plug 17 in for y and then solve for x. This will give us our estimated GPA.

17 = −5𝑥 + 29

−29 − 29

−12 = −5𝑥

÷ (−5) ÷ (−5)

2.4 = 𝑥

Scan this QR code to go to a video tutorial on line of

best fit.

31

Year 2006 2007 2008 2009 2010 2011

Average cost for one gallon

2.59 3.16 3.29 3.25 3.51 3.56

Start by entering this data into the list in your calculator. 2006 can be year 1, 2007 can be year 2, etc. Then hit the stat button again, scroll over to calc, and select number 4 (LinReg) Press enter twice and your results window will show. The line of best fit is 𝑦 = .167𝑥 + 2.64 To answer the second part of the question, we first need to determine what number the year 2017 would be associated with. Since 2011 was year 6, 2017 would be year 12. To predict the gas price in 2017, we will plug 12 in for x in our line of best fit.

𝑦 = .167 (12) + 2.64 = $4.64 Sometimes a curve (parabola) will fit the data better than a straight line will. You could quickly determine which is more appropriate by typing your data points into your calculator and graphing the points. If the points look like the picture below, a curve may be a better fit for the data. To find the curve of best fit, you follow the same procedure as above, but you select number 5 (QuadReg) instead of 4 (LinReg).

STAT ENTER

32

Example 3: What is the equation of the curve of best fit for the data below?

Time

(sec)

Height

(feet)

0 5

1 12

2 21

3 19

4 14

Start by entering this data into the list in your calculator. Time will be your L1 (x) values and Height will be L2 (y). Then, hit the stat button again, scroll over to calc, and select number 5 (QuadReg). Press enter twice and your results window will show:

The curve of best fit is 𝑦 = −2.5𝑥2 + 12.5𝑥 + 4.2

Scan this QR code to go to a video tutorial on curve of

best fit.

The question told us that we will find a CURVE of fit, so we

know we should attempt QuadReg.

STAT ENTER

This graph shows the 5 data points and the

curve of best fit.

33

Scatterplots 1. Using the data in the table, find theline of best fit. Then, use this information to

predict how long the average 9-month-old rattlesnake is.

Age (months)

1 3 5 7 10 12 15 24

Length (inches)

8 16 25 36 49 50 64 71

2. Use the data in the table to find the quadratic curve of best fit. Then, use this

information to determine at what time the ball will reach its maximum height.

Time (sec) 0 1 2 3 4 5

Height (ft) 4 8 17 21 20 14

34

Answers to the problems: Relations and Functions 1.

x y

4 4

-3 4

0 4 It is a function because none of the x-values are repeated 2. A. Yes. Passes the vertical line test B. No. Does not pass the vertical line test. 3. It is not a function 4. Yes

Domain & Range 1.Domain {-2,0,5} Range {4,3,-6} 2. A. Domain {All Real Numbers} Range {𝑦: 0 ≤ 𝑦 < ∞} B. Domain {𝑥: −∞ ≤ 𝑥 < ∞} Rangel {𝑦: −∞ ≤ 𝑦 < ∞}

Functions and Graphs 1.

2.

x y

-1 -3

0 -5

1 -3

3. B Slope 1. 𝑚 = 0

2. 𝑚 = −5

3

3. 𝑟 = −7 4. 𝑟 = 2 Slope-Intercept and Transformations 1. 𝑚 = −7 , 𝑏 = 12

2. 𝑦 = 2

5𝑥 − 4

3. 𝑦 = 𝑥 + 1 4. 𝑏 = −3

35

5. 6. B 7. 𝑦 = −2𝑥 + 10

8. 𝑦 = −1

3𝑥 − 1

9. 10.

11. 𝑦 =1

3𝑥 − 2

Direct and Inverse Variation 1. A 𝑁𝑒𝑖𝑡ℎ𝑒𝑟; B 𝐼𝑛𝑣𝑒𝑟𝑠𝑒 2. 𝑦 = 6𝑥 3. 𝑦 = 5 4. 9.4 ℎ𝑜𝑢𝑟𝑠

5. 𝑘 = 5

2

6. 6 Systems of Equations

1. ( 5

2 , −

3

2 )

2. ( −13

3 , 6 )

3. ( 5 , −2 )

4. 1380 Children

320 Adults

5. 5 dress shirts

7 t-shirts

36

5.

6. 𝑦 > 3𝑥 − 1 7.

Scatterplots

1. 𝑦 = 2.88𝑥 + 12.17

9 month old snake ≈ 38.09"

2. 𝑦 = −1.61𝑥2 + 10.61𝑥 + 2.21

Ball reaches maximum height at ≈ 3.3 𝑠𝑒𝑐𝑠