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STANDARDS OF LEARNING

CONTENT REVIEW NOTES

ALGEBRA I Part II

2nd 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. They have been revised this year as part of an internship process. 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.

Algebra I Blueprint Summary Table

Reporting Categories No. of Items SOL

Expressions & Operations 12 A.1

A.2a – c

A.3

Equations & Inequalities 18 A.4a – f

A.5a – d

A.6a – b

Functions & Statistics 20 A.7a – f

A.8

A.9

A.10

A.11

Total Number of Operational Items 50

Field-Test Items* 10

Total Number of Items 60

* These field-test items will not be used to compute the students’ scores on the test.

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4

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

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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

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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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Systems of Inequalities A.5 The student will

c) solving practical problems involving inequalities; and d) represent the solution to a system of inequalities graphically.

A linear inequality can be formed by replacing the equal sign in any linear equation

with an inequality symbol. The solutions for a linear inequality are any ordered pairs

that make it a true statement.

Example 1: Identify which ordered pairs are a solution of 𝑦 < 2𝑥 + 5 .

(0, 0) (−3, 1) (4, 4)

0 <? 2(0) + 5

0 < 5

This is true, therefore (0, 0)

is a solution.

1 <? 2(−3) + 5

1 <?− 6 + 5

1 < −1

This is false, therefore

(−3, 1) is NOT a solution.

4 <? 2(4) + 5

4 <? 8 + 5

4 < 13

This is true, therefore (4, 4)

is a solution.

As you can see from the last example, linear inequalities will have more than one solution. In fact, they will have infinitely many solutions. The graph of a linear inequality will indicate all of the solutions, and it is called a half-plane, and is bounded by a boundary line. All of the points on one side of this boundary are solutions, while all of the points on the other side of the boundary are not solutions. You graph a linear inequality the same way that you graph a linear equation. The line that you graph will either be dashed or solid depending on the inequality symbol. Dashed lines are used for < and >. This indicates that the points on the line are not part of the solution set. Solid lines are used for ≤ and ≥. This indicates that the points on the line are part of the solution set. To determine which half-plane to shade in, you can select one point that is not on the graph and determine if it is a solution or not. If it is a solution, shade that side of the boundary. If it is not a solution, shade the other side. The point (0, 0) is often an easy point to check with.

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Example 2: Graph 𝑦 <1

2𝑥 − 2

First, determine if you will be using a dashed or solid line (dashed in this case because you have < ) Then, graph the equation of the line by plotting the y-intercept and counting the slope as rise over run. Once you have a couple of points graphed, connect them with a dashed line. Finally, figure out which half-plane to color in. Select a point to see if it works.

0 <?1

2(0) − 2

0 < −2

This is false. Therefore, (0, 0) is NOT a solution. So, we will shade the other side.

All of the points on the shaded side will satisfy the inequality.

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

inequalities.

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Example 3: Graph 2𝑥 − 𝑦 ≥ 6

First, determine if you will be using a dashed or solid line (solid in this case because you have > greater than or = equal to. Before we graph this equation, we should put it in slope intercept form! Remember that if you multiply or divide by a negative number you will have to switch the inequality symbol’s direction.

2𝑥 − 𝑦 ≤ 6

−2𝑥 − 2𝑥

−𝑦 ≤ −2𝑥 + 6

÷ (−1) ÷ (−1)

𝑦 ≥ 2𝑥 − 6

Now we can put a point at the y-intercept, and count the slope as rise over run. Finally, figure out which half-plane to color in. Select a point to see if it works. You can plug this point into the original equation or the transformed equation.

2(0) − 0 ≤? 6

0 ≤ 6

This is true. Therefore, (0, 0) is a solution. So, we will shade that side.

You divided by −1

here! Don’t forget to

switch the sign!

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All of the points on the shaded side will satisfy the inequality.

To solve a system of linear inequalities, you will graph both inequalities on the same

coordinate plane. The solution is any area that is shaded for both inequalities.

Example 4: Solve the system of inequalities by graphing.

𝑦 ≤ −2𝑥 − 1

𝑦 > 𝑥 + 5

First, graph the first equation with a solid line and determine which half-plane you

should shade by choosing a point and verifying.

0 ≤?− 2(0) − 1

0 ≤ −1

Now, graph the second equation on the same graph and determine which side to shade.

0 >? 0 + 5

0 > 5

Does the point (0, 0) work?

No, so shade the other side!

Does the point (0, 0) work?

No, therefore shade the other side!

The solutions to the system of

inequalities are all of the points that are shaded

from both inequalities.

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

inequalities.

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Systems of Inequalities 1. Is (2, 12) a solution to the inequality 𝑦 < 4𝑥 + 4 ? 2. Is (−4, 0) a solution to the inequality 4𝑦 ≥ 𝑥 − 5 ? 3. Is (2, 2) a solution to this system of inequalities? 𝑦 > 𝑥 − 4 𝑦 < 4𝑥 − 5 4. Graph 3𝑦 > 2𝑥 − 9 . 5. Graph 𝑥 + 2𝑦 ≤ 8 . 6. Write the inequality that is graphed here

7. Solve the system of inequalities by graphing.

𝑦 > 2𝑥 − 3 𝑦 < −𝑥 − 4

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Laws of Exponents & Polynomial Operations A.2 The student will perform operations on polynomials, including

a) applying the laws of exponents to perform operations on expressions;

Monomial is a single term. It could refer to a number, a variable, or a product of a number and one or more variables. Some examples of monomials include:

14𝑎𝑏² − 6𝑑 1 1

2𝑥²𝑦𝑧²

When you multiply monomials that have a common base, you add the exponents.

Example 1: Multiply 𝑏2 ∙ 𝑏5

𝑏2 ∙ 𝑏5 = 𝑏2+5 = 𝑏7

This works because when you raise a number or variable to a power, it is like multiplying it by itself that many times. When you then multiply this by another power of the same number or variable, you are just multiplying it by itself that many more times.

Example 2: 𝑅𝑒𝑤𝑟𝑖𝑡𝑒 43 𝑎𝑛𝑑 42𝑎𝑠 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝑝𝑟𝑜𝑏𝑙𝑒𝑚𝑠. 𝑈𝑠𝑒 𝑡ℎ𝑖𝑠 𝑡𝑜 𝑠𝑖𝑚𝑝𝑙𝑖𝑓𝑦 43 ∙ 42.

43 = 4 ∙ 4 ∙ 4 42 = 4 ∙ 4

43 ∙ 42 = 4 ∙ 4 ∙ 4 ∙ 4 ∙ 4 = 45

Example 3: Simplify 1

2𝑥𝑦3𝑧5 ∙ 14𝑥3𝑧 ∙ 𝑦6𝑧2

(1

2 ∙ 14) (𝑥 ∙ 𝑥3)(𝑦3 ∙ 𝑦6)(𝑧5 ∙ 𝑧 ∙ 𝑧2)

7𝑥4𝑦9𝑧8 When you raise a power to a power, you multiply the exponents.

(32)4

32 ∙ 32 ∙ 32 ∙ 32

32+2+2+2

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Example 4: Simplify (𝑎2𝑏)4

(𝑎2𝑏)4 = 𝑎2∙4𝑏1∙4

𝑎8𝑏4

This means 3² times itself 4 times!

Scan this QR code to go to a video tutorial on

multiplying monomials!

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Example 5: Simplify (5𝑑5𝑒7𝑓2)3

53𝑑5∙3𝑒7∙3𝑓2∙3

125𝑑15𝑒21𝑓6

Often, you will be asked to multiply monomials and raise powers to a power. Make sure that you follow the ORDER OF OPERATIONS! Raise to powers first, then multiply.

Example 6: Simplify (−2𝑥³𝑦𝑧²)³(2𝑥³𝑦𝑧²)⁴

(−8𝑥9𝑦3𝑧6) ∙ (16𝑥12𝑦4𝑧8)

−128𝑥21𝑦7𝑧14

Laws of Exponents Simplify each expression

1. 6𝑎3𝑏5(−𝑎𝑏2)

2. (−5𝑥2𝑦𝑧3)2

3. [(𝑚𝑛6𝑝3)2]4

4. (𝑎4𝑏2𝑐7)4 (13𝑎2𝑐)3 When you divide monomials with like bases, you will subtract the exponents.

Anything raised to the zero power is equal to ONE!

𝑥0 = 1 150 = 1 (−235𝑦7)0 = 1 To find the power of a quotient, raise both the numerator and the denominator to the power. (Remember to follow the order of operations!)

(𝑎

𝑏)

5

= 𝑎5

𝑏5

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Example 7: 𝑛5𝑝7

𝑛4𝑝

𝑛5𝑝7

𝑛4𝑝= 𝑛5−4 𝑝7−1 = 𝑛 𝑝6

Example 8: (𝑎2𝑏𝑐5

𝑎2𝑐2 )3

(𝑎2𝑏𝑐5

𝑎2𝑐2)

3

= (𝑎2−2𝑏𝑐5−2)3 = (𝑎0𝑏𝑐3)3 = 13𝑏3𝑐9 = 𝑏3𝑐9

You will also see negative exponents in monomials. When you have a negative exponent, you will reciprocate that variable (move it to the other side of the fraction bar) and the exponent will become positive.

As an example: 2−3 = .125 = 1

8 𝑜𝑟 2−3 =

1

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When simplifying monomials with negative exponents, you can start by ‘flipping over’ all of the negative exponents to make them positive. Then, simplify.

Example 9: 𝑎−3𝑏2𝑐−5

𝑎−7𝑏𝑐10

𝑎−3𝑏2𝑐−5

𝑎−7𝑏𝑐10

𝑎7𝑏2

𝑎3𝑏𝑐10𝑐5

𝑎4𝑏

𝑐15

Example 10: (2𝑥2𝑦−4

3𝑥𝑦5 )−3

(2𝑥2𝑦−4

3𝑥𝑦5)

−3

= (3𝑥𝑦5

2𝑥2𝑦−4)

3

= (3𝑥𝑦5𝑦4

2𝑥2)

3

= (3𝑦9

2𝑥)

3

= 27𝑦27

8𝑥3

Remember that anything to the zero power equals 1!

Scan this QR code to go to a video tutorial on

dividing monomials!

Scan this QR code to go to a video tutorial on

simplifying monomials

with negative exponents!

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Exponents Laws of Exponents

Simplify each expression

5. 33𝑎5𝑏9

11𝑎6𝑏2

6. (𝑥4𝑦2

2𝑥𝑦)

3

7. (2𝑥𝑦−3)5

8. 𝑛4𝑛2𝑚−5

𝑛6𝑚−2𝑝0

9. 15𝑎𝑏−2𝑐5

9𝑎−4𝑏2𝑐2

10. (6𝑥4𝑦2𝑧−3

9𝑥−2𝑦0𝑧−1 )

−2

Polynomials A.2 The student will perform operations on polynomials, including

b) adding, subtracting, and multiplying polynomials.

Adding and subtracting polynomials is the same as COMBINING LIKE TERMS. In order for two terms to be like terms, they must have the same variables and the same exponents.

Like Terms NOT Like Terms

5𝑎𝑏2, −3𝑎𝑏2,2

3𝑎𝑏2 5𝑎𝑏2, −3𝑎2𝑏,

2

3𝑎𝑏

Each of these terms contain an ‘𝑎𝑏2 ‘, therefore they are like terms.

Although these terms have the same variables, corresponding variables do not

have the same exponents. Therefore, these are NOT like terms.

Example 1: (2𝑥2𝑦 + 5𝑥𝑦 − 7𝑦2) + (4𝑥2𝑦 − 10𝑥𝑦 + 3𝑦2)

(2𝑥2𝑦 + 4𝑥2𝑦) + (5𝑥𝑦 − 10𝑥𝑦) + (−7𝑦2 + 3𝑦2)

6𝑥2𝑦 − 5𝑥𝑦 − 4𝑦2

Like terms are underlined here. Remember that each term takes the sign in front of it!

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Remember that if you are subtracting a polynomial, you are subtracting all of the terms (Therefore, you must distribute the negative to each term first!)

Example 2: (−3𝑎𝑏 − 5𝑎4𝑏2 + 𝑏) − (4𝑎𝑏 − 6𝑏)

−3𝑎𝑏 − 5𝑎4𝑏2 + 𝑏 − 4𝑎𝑏 + 6𝑏

−7𝑎𝑏 − 5𝑎4𝑏2 + 7𝑏

Polynomials Simplify each expression

1. (𝑚𝑛𝑝2 − 7𝑛𝑝2 + 12𝑚𝑛) + (4𝑚𝑛𝑝2 − 3𝑚𝑛)

2. (4𝑎 + 9𝑏 − 3𝑐 + 2𝑑) + (2𝑎 − 𝑏 − 5𝑐 + 3𝑑2)

3. (12𝑥2 − 6𝑥𝑦 + 9𝑦2) − (3𝑥2 + 𝑥𝑦 − 4𝑦2)

4. (32𝑎𝑏2 + 5𝑎2𝑏 − 21𝑏2) − (𝑎𝑏 + 14𝑏2 − 5𝑎2𝑏) 5. (2𝑡𝑢 − 8𝑢 + 7𝑡) + (−𝑡𝑢 − 4𝑡) − (3𝑢 + 𝑡 − 5𝑡𝑢) To multiply a polynomial by a monomial, simply distribute the monomial to each term in the polynomial. You will use the rules of exponents to simplify each term.

Example 3: 5𝑥 (3𝑥2 − 6𝑥𝑦 + 2𝑦2)

(5𝑥 ∙ 3𝑥2) − (5𝑥 ∙ 6𝑥𝑦) + (5𝑥 ∙ 2𝑦2)

15𝑥3 − 30𝑥2𝑦 + 10𝑥𝑦2

Example 4: −2𝑎2𝑏 (5𝑎𝑏3 − 6𝑎2𝑏5 + 𝑎2𝑏 − 𝑎𝑏3 ) (−2𝑎2𝑏 ∙ 5𝑎𝑏3) + (−2𝑎2𝑏 ∙ −6𝑎2𝑏5 ) + (−2𝑎2𝑏 ∙ 𝑎2𝑏 ) + (−2𝑎2𝑏 ∙ −𝑎𝑏3)

−10𝑎3𝑏4 + 12𝑎4𝑏6 − 2𝑎4𝑏2 + 2𝑎3𝑏4

−8𝑎3𝑏4 + 12𝑎4𝑏6 − 2𝑎4𝑏2

Distribute the negative to everything in the second set of parentheses!

Then, COMBINE LIKE TERMS!

Scan this QR code to go to a video tutorial on

adding and subtracting polynomials.

Distribute the 5𝑥 to each term.

Then, simplify each term

Don’t forget to check for like terms!

Scan this QR code to go to a video tutorial on

multiplying monomials

and polynomials.

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To multiply two polynomials together, distribute each term in the first polynomial to each term in the second polynomial. When you are multiplying two binomials together this may be called FOIL. FOIL stands for:

F – First – multiply the first term in each binomial together O – Outer – multiply the outermost term in each binomial together I – Inner – multiply the innermost term in each binomial together L – Last – multiply the last term in each binomial together (This is the exact same as distributing the first term, then distributing the second term) Don’t forget to combine like terms when possible.

Example 5: (2𝑥 + 5)(3𝑥 − 2) First Outer Inner Last

(2𝑥 ∙ 3𝑥) + (2𝑥 ∙ −2) + (5 ∙ 3𝑥) + (5 ∙ −2)

6𝑥2 − 4𝑥 + 15𝑥 − 10

6𝑥2 + 11𝑥 − 10

Example 6: (𝑎2 + 2𝑏2)(4𝑎 − 3𝑎𝑏 + 6𝑏)

(𝑎2)(4𝑎 − 3𝑎𝑏 + 6𝑏) + (2𝑏2)(4𝑎 − 3𝑎𝑏 + 6𝑏)

4𝑎3 − 3𝑎3𝑏 + 6𝑎2𝑏 + 8𝑎𝑏2 − 6𝑎𝑏3 + 12𝑏3

Example 7: (4𝑦 − 3)2 (4𝑦 − 3)(4𝑦 − 3) (4𝑦 ∙ 4𝑦) + (4𝑦 ∙ −3) + (−3 ∙ 4𝑦) + (−3 ∙ −3)

16𝑦2 − 12𝑦 − 12𝑦 + 9

16𝑦2 − 24𝑦 + 9

Polynomials

6. −2𝑚𝑛2(4𝑚2𝑛 − 3𝑚𝑛) 7. 2𝑎𝑏2𝑐 (4𝑎 + 𝑏 − 3𝑐) + 5𝑎𝑏2𝑐 (6𝑎 − 3𝑏 − 5𝑐) 8. (6𝑥 + 5)(6𝑥 − 5)

9. (5𝑎2𝑏 − 2𝑏2)(4𝑎𝑏 + 𝑏)

10. (3𝑥𝑦 − 5𝑥)2

Remember that to square something means to multiply it by itself!

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Answers to the problems: 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

Systems of Inequalities 1. No

2. Yes

3. Yes

4.

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5.

6. 𝑦 > 3𝑥 − 1 7.

24

Laws of Exponents

1. −6𝑎4𝑏7

2. 25𝑥4𝑦2𝑧6

3. 𝑚8𝑛48𝑝24

4. 2197𝑎22𝑏8𝑐31

5. 3𝑏7

𝑎

6. 𝑥9𝑦3

8

7. 32𝑥5

𝑦15

8. 1

𝑚3

9. 5𝑎5𝑐3

3𝑏4

10. 9𝑧4

4𝑥12𝑦4

Polynomials

1. 5𝑚𝑛𝑝2 − 7𝑛𝑝2 + 9𝑚𝑛

2. 6𝑎 + 8𝑏 − 8𝑐 + 2𝑑 + 3𝑑2

3. 9𝑥2 − 7𝑥𝑦 + 13𝑦2

4. 32𝑎𝑏2 + 10𝑎2𝑏 − 35𝑏2 − 𝑎𝑏 5. 6𝑡𝑢 − 11𝑢 + 2𝑡

6. −8𝑚3𝑛3 + 6𝑚2𝑛3

7. 38𝑎2𝑏2𝑐 − 13𝑎𝑏3𝑐 − 31𝑎𝑏2𝑐2

8. 36𝑥2 − 25

9. 20𝑎3𝑏2 + 5𝑎2𝑏2 − 8𝑎𝑏3 − 2𝑏3

10. 9𝑥2𝑦2 − 30𝑥2𝑦 + 25𝑥2