summer review packet for algebra 2 cp/honors review packet for algebra 2 cp/honors name current...

Summer Review Packet For Algebra 2 CP/Honors

Name Current Course Math Teacher Introduction

Algebra 2 builds on topics studied from both Algebra 1 and Geometry. Certain topics are sufficiently involved that they call for some review during the year in Algebra 2. However, some topics and basic skills are so fundamental to any Algebra 1 course that MASTERY of these topics is expected and required PRIOR to beginning Algebra 2. This packet covers these topics, which are:

I. Solving equations in one variable. p.1

II. Solving inequalities in one variable. p.2 III. Linear Equations

A. Graphing a linear equation. p.3 B. Finding the slope. p.3 C. Writing a linear equation. p.3 D. Writing a linear equation of parallel and perpendicular lines. p.4 IV. Solving a system of linear equations. A. Graphing method. p.5-6 B. Substitution method. p.7 C. Elimination method. p.8 V. Graphing a system of linear inequalities. p.9

VI. Factoring p.10 VII. Simplifying Radicals p.11 VIII. Simplifying Expressions with Exponents p.12 IX. Solving Absolute Value Equations and Inequalities p. 12 Practice exercises begin on p.13 Answers to practice exercises begin on p.21 We will be using a Ti-83 or Ti-84 in our classes. Note about the calculators: The school will issue each student who needs/wants a Ti-83 or Ti-84 graphing calculator, just as we issue a textbook to each student. If it is lost or broken, the student must pay the replacement value of about $90. A student may prefer to purchase his/her own. If so, we recommend the Ti-84 Plus (Silver Edition optional). We then highly recommend the calculator be etched with the student’s name clearly on the front face, and on the back for security purposes. A sticker or a name written with a marker is not enough.

During this summer you are to complete this review packet, following the directions for each section. Although you will have some opportunity to get some help, if you need a lot of help, you should consider getting it before school begins. You will be tested on this material shortly after the beginning of school. The topics on computation with and without a calculator will require true mastery – a 100% result, and will be formative up to that point. We recommend you do this packet close to the onset of school.

DO NOT LOSE THIS PACKET!! It will also be available from the school website via the math department.

Algebra 2 Summer Review Packet

Solved Examples I. Solving equations in one variable.

Example 1: Solve 92x31

=+

Solution:

Example 2: Solve ( ) 1178y3 =−+ Solution:

Example 3: Solve ( ) a8a262a7 ++=−−

Solution:

( )

7 a 4

28 4a4

20 20 8 20 a4 a3 a3

8a3 20 a7 a8a2614a7a8a262a7

=

=

++=−−−

+=−++=−−++=−−

Example 4: Solve )1n(2n5)n1(3 +=+−

Solution:

or solution no 23

n2 n2 2n2 n2 3 2n2 n5 n3 3

)1n(2 n5)n1(3

∴=−−

+=++=+−

+=+−

Eliminate the +2 by subtracting 2 from both sides of the equation. Eliminate the by multiplying by its reciprocal, 3, on both sides.

Use the distributive property and combine to simplify the left side of the equation. Eliminate the +17 by subtracting 17 from both sides. Eliminate the 3 by dividing both sides by 3.

Use the distributive property and combine to simplify both sides of the equation. Eliminate the variable term on the right by subtracting 3a. Eliminate the –20 by adding 20 to both sides. Eliminate the 4 by dividing both sides by 4.

Continue as in the previous example. Correct steps have resulted in a false statement. Conclude that the equation has no solution.

Ø or { }

LHS 1


Solved Examples II. Solving inequalities in one variable. Example 5: Solve x6233x4 +≥+−

Solution:

{ }2xx

2x 10 1020 x10

3 3 233x10

x6 x6 x6233x4

−≤

−≤−−≥−−−≥+−

−−+≥+−

Example 6: Solve ( )4x8413x

52

+−<+

Solution:

( )

−<

−<

−<

−−

<+

++

+−<+

+−<+

65xx

65 x

2 x5

123 3

13x5

12x2 x2

1x23x52

4x8413x

52

Example 7: Solve ( )8x2x5110x3 −−+>+

Solution:

( )

{ }numbers real all is

910 3 3

93103 8251103 82x5x110x3

xx

xxxx

xxx

∴

>−−

+>++−+>+−−+>+

Subtract 6x from both sides of the inequality. Subtract 3 from both sides of the inequality. Divide both sides of the inequality by –10. Remember to reverse the inequality symbol when multiplying or dividing by a negative number.

Distribute .

Add 2x to both sides of the inequality. Subtract 3 from both sides of the inequality.

Multiply both sides of the inequality by .

Use the distributive property and combine to simplify both sides of the inequality. Subtract 3x from both sides of the inequality. Correct steps have resulted in a true statement. Conclude that the solution is all real numbers.

LHS 2


Solved Examples III. Linear Equations A. Graphing a linear equation Example 8: Find the slope and the y-intercept of the line whose equation is 3y = 2x + 9 and use them to graph

the equation. Solution:

The slope is the coefficient of the x-term, 32 , and the y-intercept is the constant

term, 3. To graph the line, plot the y-intercept, (0 , 3). Then, since the slope is 32 , move 2

units up(rise) and 3 units right (run) to locate a second point. (See below.)

III. Linear Equations B. Finding the slope. C. Writing a linear equation. Example 9: a) Find the slope of the line passing through the points (-2 , 5) and (4 , 8).

b) Given that the y-intercept is 6, write an equation for the line.

Solution: a) 21

63

2458

xxyyslope

12

12 ==+−

=−−

=

b) In a linear equation of the form bmxy += , the slope is the coefficient of the x-term or linear term, namely m. The numerical term or constant term is the y-intercept, namely b. So, starting with bmxy += , substitute ½ for m and 6 for b.

Therefore, the equation is 6x21y +=

Divide by 3 to solve for y. Now the equation is in the familiar form .

+3 run

+2 rise

LHS 3


Solved Examples III. Linear Equations D. Writing a linear equation of parallel and perpendicular lines.

Example 10: a) Write an equation of a line in slope-intercept form parallel to 3x32y −= and passes through the

point (3, –2)

b) Write an equation of a line in point-slope form perpendicular to 3x32y −= and passes through

the point (3, –2)

Solution: a) The32slope = since parallel lines have the same slope. Use the values for slope and the given

coordinate, and substitute the values for m, x, and y in the equation bmxy += to solve for b.

( )

b4b for solve to 2 subtract 2 2

b22

b3322

bmxy

=−−−

+=−

+=−

+=

Therefore the equation parallel to 3x32y −= and passes through the point (3, –2) is:

4x32y −=

b) The23slope −= since perpendicular lines have opposite reciprocal slopes. The point-slope form is

much easier to use. It requires substituting in the values for the slope and the point.

The form is ( )11 xxmyy −=− . Substitute m with 23

− and the 1x with 3 and 1y with –2 taken from

the given coordinate (3, –2).

Therefore the equation in point-slope form of the line perpendicular to 3x32y −= and passes through

the point (3, -2) is:

( )3232 −−=+ xy

LHS 4


Solved Examples IV. Solving a system of linear equations. A. Graphing method. Example 11: Solve the following system by graphing.

4x2y

1xy 31

−=

+=

Solution: To solve the system of equations by graphing, you need to graph both equations carefully and then locate

the point of intersection. The point of intersection is the solution to the system. Since both equations are already in slope-intercept form it is easy to extract the slope and y-intercept of

each equation and use that information to graph each accordingly as shown previously in the section on page 3 called III. Linear Equations A. Graphing a linear equation.

The solution (the point of intersection) is (3 , 2)

Example 12: Solve the following system by graphing.

4y4x36yx=−

=+

Solution: Put each equation into slope-intercept form. The solution is (4 , 2)

LHS 5


Solved Example IV. Solving a system of linear equations. A. Graphing method. Example 13: Solve the following system by graphing.

11y2xy76x3

=+=+

Solution: Put each equation into slop-intercept form.

The y-intercepts for the two equations are not nice whole number, thus making it difficult to graph those points. Now make a table of x and y values and find nice points to graph. Only two points are needed to graph a line.

The solution is (5 , 3)

x y

-2 0 -1 0.42

0 0.85

1 1.29

2 1.71

3 2.14

4 2.57

5 3

x y

-2 6.5

-1 6 0 5.5

1 5

Use these two points to graph

Use these two points to graph

LHS 6


Solved Example IV. Solving a system of linear equations. B. Substitution method. Example 14: Solve the following system by using substitution.

27y2x313y3x=+−

=+

Solution: To solve the system of equations by substitution, you need to solve for x or y with one of the equations.

Your choice here can save time. Let’s choose to solve the 1st equation for x. This will require only 1 step.

Substitute the equivalent expression for x into the other equation. Continue to solve for y accordingly. At this point substitute the solution for y into either of the two original equations. Again your choice here can save time. Let’s choose equation1. The solution is (-5 , 6)

LHS 7


Solved Example IV. Solving a system of linear equations. C. Elimination method. Example 15: Solve the following system by using substitution.

29y3x24y5x3

=−−=+

Solution: To solve the system of equations by elimination, you need to write equivalent equations containing the

same coefficient for either x or y. Multiply the first equation by 3 and the second by 5. Then the variable y can be eliminated by adding the two equations.

Find y by substituting 7 for x in either equation. Let’s use the 1st one. The solution is (7 , –5)

Multiply by 3 Multiply by 5

LHS 8


Solved Examples V. Graphing a system of linear inequalities. Graphing a Linear Inequality Slope- intercept form: y < mx + b 1st graph the y-intercept 2nd use the slope to graph one or more points.

Connect points to make a solid or dashed line. Shade above or below

The solution to a system of inequalities is the common area that is shaded. Example 16: Graph the solution to Example 17: Graph the solution to

2

42<

−≥y

xy

3x

21

y

4x

+−>

−≥

line shading

solid dashed above below

≤ ● ●

≥ ● ●

< ● ●

> ● ●

Special lines: Horizontal lines Vertical lines Shade in above or below Shade in left or right side y > 2 y ≥ 2 y < 2 y ≤ 2 x > 2 x ≥ 2 x < 2 x ≤ 2

shading above above below below right right left left line dashed solid dashed solid dashed solid dashed solid

LHS 9


Solved Examples VI. Factoring Factoring (Common Monomial Factor) Example 18: Factor 2223 2015 yzxzyx − Solution: Find the greatest common factor on the terms in the polynomial. GCF is yzx 25 . Divide each term in the polynomial by the GCF to find the other factor.

xyyzx

zyx 35

152

23= and z

yzxyzx 4

520

2

22=

( )zxyyzxyzxzyx 4352015 22223 −=−

Factoring (Difference of Squares) Example 19: Factor 2516 2 −x Solution: A term is a square if the exponents on all variables in it are even and the coefficient is the square of an

integer. Both 16x2 and 25 are squares. This will only factor if the two squares are subtracted – thus a difference of squares. In particular, 16x2 = (4x)2 and 25 = (5)2. The binomial factors as the sum and the difference of the squares roots:

16x2 – 25 = (4x + 5)(4x – 5)

Factoring (Perfect Square Trinomials) Example 20: Factor 4x2 – 20xy + 25y2 Solution: A perfect square trinomial is a polynomial with three terms that results from

squaring a binomial. In other words, perfect square trinomial = (some binomial)2. To factor, we must first check that the trinomial is a perfect square by answering the following three questions:

1. Is the first term a square? Answer: Yes, 4x2 = (2x)2 2. Is the last term a square? Answer: Yes, 25y2 = (5y)2

3. Ignoring the sign at this time, is the middle term twice the product of 2x and 5y? Answer: Yes, 20xy = 2 (2x⋅5y)

Having established that the trinomial is a perfect square, it is easy to find the binomial of which it is a square; its terms are in the parenthesis above.

4x2 – 20xy + 25y2 = (2x - 5y)2 Since the middle term in the trinomial is negative, the factored result is a subtraction.

Factoring (Product – Sum) Example 21: Factor y2 + 14y + 40 Solution: If, as above, the trinomial is in standard form, the product number is the numerical term, 40, and the sum number is the coefficient of the linear term, 14. The unique

pair of number whose sum is 14 and product is 40 is 4 and 10. Therefore, the trinomial factors as y2 + 14y + 40 = (y + 4)(y + 10)

(Note: all factoring problems can be checked by multiplying out)

LHS 10


Solved Examples VII. Simplifying Radicals HERE ARE THE FIRST TWENTY square numbers and their roots: Square numbers 1 4 9 16 25 36 49 64 81 100 121 144 169 196 225 256 289 324 361 400

Square roots 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 We write, for example, 525 = . "The square root of 25 is 5." This mark is called the radical sign (after the Latin radix = root). The number under the radical sign is called the radicand. In the example, 25 is the radicand.

Simplify the following radical expressions.

Example 22: =45 Example 23: =4

32

Solution 535959 =⋅=⋅ Solution 22242484

32=⋅=⋅==

or 222

242

2162

216232

432

==⋅

=⋅

==

Rationalizing the denominator

Example 24: =2

3 Example 25: =+ 312

Solution 2

23423

22

23

==⋅ Solution ( ) ( ) 312

31231

3123131

312

−−=−+

=−+

=+

+⋅

−

Properties of Square Roots:

Product Property

The square root of a product is equal to the product of each square root

; when and

Quotient Property

The square root of a quotient is equal to the quotient of each square root.

; when and

LHS 11


Solved Examples VIII. Simplifying Expressions with Exponents

Integral Exponents times n ....bbbbbn ⋅⋅⋅=

b: base n: exponent

273333

93333

2

=⋅⋅=

=⋅= 273333

93333

2

−=⋅⋅−=−

−=⋅−=− ( ) ( )( ) ( ) ( ) 27333)3(

933)3(3

2

−=−⋅−⋅−=−

=−⋅−=−

Rules of Exponents Product of Powers

Quotient of Powers Power of a Power Power of a Product

nmnm bbb += nmn

mb

bb −= ( ) nmnm bb ⋅= ( ) mmm baab =

Power of a Quotient

Zero Exponent Negative Exponent

m

mm

ba

ba

=

10 =b nn

bb 1

=− nn

ab

ba

=

−

Simplify each expression.

Example 26: ( )( ) =−− 43 28 xyx Example 27: =−2

6

424

cc

Solution ( ) ( )

Rule Powers of Productapply 16x

product the rearrange 287

43

y

yxx

=

⋅⋅⋅−⋅− Solution

Rule Powers of Quotientapply 6c

fraction reduce 16

8

2

6

=⋅

⋅−c

c

Example 28: =2

53

4)10)(2(

aaa Example 29: =

abcbca

32)(128 324

Solution 62

8

1

535

420

4102 a

aa

aaa

==⋅⋅⋅

− Solution 523

63444 cba

abccba

=

Example 30: =

−−

−− 0

1527

647

1442

zyxzyx

Solution = 1 IX. Absolute Value Equations and Inequalities For this section, go to the link http://www.khanacademy.org/math/algebra/solving-linear-equations-and-inequalities/absolute-value-equations/v/absolute-value-equations and watch the videos on absolute value equations and inequalities.

LHS 12

http://www.khanacademy.org/math/algebra/solving-linear-equations-and-inequalities/absolute-value-equations/v/absolute-value-equations

http://www.khanacademy.org/math/algebra/solving-linear-equations-and-inequalities/absolute-value-equations/v/absolute-value-equations


Exercises I. Solving equations in one variable. Solve each equation.

1. 5)2v(431 −=+−

2. ( ) ( ) 43cc21c −=−+−− 3. 1n51n5 −=+

4. c57

c1123=

− 5. ( ) 2y5131y34 +=+− 6. ( ) ( )a232142a4 −−=+

II. Solving inequalities in one variable. Solve each inequality. 1. ( ) ( )4b25b2b3 +>−−

2. 18x7

81x7

+>+ 3. ( )

101x5

53x3 −<

−

4. 3x1

48x

>−+ 5. w2

4w

5120 −≥

−

LHS 13


Exercises

III. Linear Equations B. Finding the slope. C. Writing a linear equation.

Find the slope of the line passing through each pair of points.

Write an equation of the line in A) slope-intercept form and B) point-slope form with the given information.

4. slope = 3 , y-intercept = 7 5. slope =

32 , (3 , 4)

6. (2 , 4) , (-2 , -16)

1. (0 , 7) and (1 , 9) 2. (8 , 1) and (-8 , 1) 3. (-2 , -6) and (-2 , 4)

LHS 14


Exercises

III. Linear Equations D. Writing a linear equation of parallel and perpendicular lines.

Write an equation of a line in slope-intercept form that is parallel to the graph of each equation and passes through

the given point.

7. 1yx4 =+ ; (–2 , 1) 8. 3y = ; (2 , 6)

Write an equation of a line in slope-intercept form that is perpendicular to the graph of each equation and passes

through the given point.

9. 3y5x2 =− ; (–2 , 7) 10. 6x = ; (4 , 2)

LHS 15


Exercises IV. Solving a system of linear equations. A. Graphing method. Graph each system of equations and state its solution. 1. 2. 3. 4.

LHS 16


Exercises IV. Solving a system of linear equations. B. Substitution method. C. Elimination method. Solve each system of equations by using substitution.

5. 17y3x

2y3x2−==+

6. 731834

=+=−

yxyx

Solve each system of equations by using elimination.

7. 15y3x727y6x5

−=+−=−

8.

2yx32

7y21x

31

−=−

=+

LHS 17


Exercises V. Graphing a system of linear inequalities.

1. 3y

y2x3−>−≤

2. 4x2y

1x21

y

−−≤

+≥

3. 13

622<+

−>+yx

yx 4.

5y5x−<

>

LHS 18


Exercises VI. Factoring.

1. 15a – 25b + 20 2. 48a3b2 + 72a2b3 3. 9a2 - 100

4. x2 + 5x + 4 5. a2 + 10ab + 24b2 6. 49a2 + 28ab + 4b2

VII. Simplifying Radicals.

1. 98 2. 60 3. 553 ⋅

4. 87 5. 3

75 6. 23

4+

VIII. Simplifying Expressions with Exponents.

1. ab

ba5

45 24 2.

abba

52 72

− 3.

yxy3

21

22−

4. 0

2

4

5.02

−

−

xyba 5.

− 23

2 23 zy

xzyx 6. ( )( )

( )( )321

225

824

yxxzyxz

−

−−

LHS 19


IX. Absolute Value Equations and Inequalities. 1. |−4𝑥| = 32 2. |2𝑥 − 3| = −1 3. 2|3𝑥 − 2| = 14

4. |𝑥 − 1| = 5𝑥 + 10 5. 14

|4𝑥 + 7| = 8𝑥 + 16 6. −2|𝑥| − 3 = 4𝑥 − 1

7. |𝑥 + 3| > 9 8. |3𝑥 − 6| + 3 < 15 9. 3|2𝑥 − 1| ≥ 21

10. 14

|𝑥 − 3| + 2 < 1 11. −2|𝑥 + 4| < 22 12. �𝑥−32� + 2 < 6

LHS 20


Answers to Exercises

I. Solving equations in one variable. p.13 II. Solving inequalities in one variable. p.13 III. Linear Equations p.14 B. Finding the slope. C. Writing a linear equation. III. Linear Equations p.15 D. Writing a linear equation of parallel and perpendicular lines. IV. Solving a system of linear equations. p.16 A. Graphing method. 1. solution is ( )2,3 − 2. solution is ( )5,4−

1. v = 6 2. c = 0 3. no solution 4. 5. y = -1 6. All Real Numbers

1. 2. no solution 3. 4. 5.

1.

2.

3. undefined slope

4. A) B)

5. A)

B)

6. A) B) or

7. 8. 9. 10.

LHS 21



IV. Solving a system of linear equations. p.16 A. Graphing method. 3. solution is ( )3,1− 4. no solution IV. Solving a system of linear equations. p.17 B. Substitution method. C. Elimination method. V. Graphing a system of linear inequalities. p.18

1. 2.

5. 6. 7. 8.

LHS 22



V. Graphing a system of linear inequalities. p.18 3. 4. VI. Factoring. p.19

1. 5 (3a – 5b + 4) 2. 24a2b2 (2a + 3b) 3. (3a + 10)(3a – 10)

4. (x + 1)(x + 4) 5. (a + 6b)(a + 4b) 6. (7a + 2b)2

VII. Simplifying Radicals. p.19

1. 27 2. 152 3. 15 4. 214 5. 5 6. 7

2412 +

VIII. Simplifying Expressions with Exponents. p.19

1. ba39 2. 5

2 6ab− 3.

16xy 4. 1 5.

3

532z

yx 6. 5

33yz

IX. Absolute Value Equations and Inequalities. p. 20

1. -8, 8 2. No solution 3. 3, −𝟓𝟑 4. −𝟑

𝟐 5. −𝟓𝟕

𝟐𝟖,−𝟕𝟏

𝟑𝟔 6. −𝟏,−𝟏

𝟑 7. x < −12 or x > 6

8. −𝟐 < 𝒙 < 𝟔 9. 𝒙 ≤ −𝟑 𝒐𝒓 𝒙 ≥ 𝟒 10. 𝑵𝒐 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 11. All real numbers are solutions

12. −𝟓 < 𝒙 < 𝟏𝟏

LHS 23

summer review packet for algebra 2 cp/honors review packet for algebra 2 cp/honors name current...

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