Summer Review Packet For Algebra 2 CP/Honors...Algebra 2 builds on topics studied from both Algebra...

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 $90 for the TI-83 and $105 for the TI-84. 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 in class when school begins, 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

LHS 1


LHS 2

Solved Examples I. Solving equations in one variable.

Example 1: Solve 92x3

1

Solution:

Example 2: Solve 1178y3

Solution:

Example 3: Solve a8a262a7

Solution:

7 a

4

28

4

a4

20 20

8 20 a4

a3 a3

8a3 20 a7

a8a2614a7

a8a262a7

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 3

Solved Examples II. Solving inequalities in one variable.

Example 5: Solve x6233x4

Solution:

2xx

2x

10 10

20 x10

3 3

233x10

x6 x6

x6233x4

Example 6: Solve 4x84

13x

5

2

Solution:

6

5xx

6

5 x

2 x5

12

3 3

13x5

12

x2 x2

1x23x5

2

4x84

13x

5

2

Example 7: Solve 8x2x5110x3

Solution:

numbers real all is

910

3 3

93103

8251103

82x5x110x3

xx

xx

xx

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 4

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, 3

2, and the y-intercept is the constant

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

2, 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) 2

1

6

3

24

58

xx

yyslope

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 6x2

1y

Divide by 3 to solve for y.

Now the equation is in the familiar form .

+3 run

+2 rise


LHS 5

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 3x3

2y and passes through the

point (3, –2)

b) Write an equation of a line in point-slope form perpendicular to 3x3

2y and passes through

the point (3, –2)

Solution: a) The3

2slope 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.

b4

b for solve to 2 subtract 2 2

b22

b33

22

bmxy

Therefore the equation parallel to 3x3

2y and passes through the point (3, –2) is:

4x3

2y

b) The2

3slope 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 2

3 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 3x3

2y and passes through

the point (3, -2) is:

32

32 xy


LHS 6

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

4x2y

1xy31

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.

4y4x3

6yx

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


LHS 7

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

11y2x

y76x3

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 8

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

27y2x3

13y3x

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 9

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

29y3x2

4y5x3

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 1

st one.

The solution is (7 , –5)

Multiply by 3 Multiply by 5


LHS 10

Solved Examples V. Graphing a system of linear inequalities. Graphing a Linear Inequality Slope- intercept form: y < mx + b 1

st 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

3x2

1y

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 11

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

zyx3

5

152

23

and zyzx

yzx4

5

202

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: 16x

2 – 25 = (4x + 5)(4x – 5)

Factoring (Perfect Square Trinomials) Example 20: Factor 4x

2 – 20xy + 25y

2

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 (2x5y) 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. 4x

2 – 20xy + 25y

2 = (2x - 5y)

2

Since the middle term in the trinomial is negative, the factored result is a subtraction.

Factoring (Product – Sum) Example 21: Factor y

2 + 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 y

2 + 14y + 40 = (y + 4)(y + 10)

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


LHS 12

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

24

2

216

2

216

2

32

4

32

Rationalizing the denominator

Example 24: 2

3 Example 25:

31

2

Solution 2

23

4

23

2

2

2

3 Solution

31

2

312

31

312

31

31

31

2

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 13

Solved Examples VIII. Simplifying Expressions with Exponents

Integral Exponents

times n ....bbbbbn

b: base n: exponent

273333

9333

3

2

273333

9333

3

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 nm

n

m

bb

b nmnm bb mmmbaab

Power of a Quotient

Zero Exponent Negative Exponent

m

mm

b

a

b

a

10 b n

n

bb

1

nn

a

b

b

a

Simplify each expression.

Example 26: 43 28 xyx Example 27: 2

6

4

24

c

c

Solution

Rule Powers of Productapply 16x

product the rearrange 28

7

43

y

yxx

Solution

Rule Powers of Quotientapply 6c

fraction reduce 1

6

8

2

6

c

c

Example 28: 2

53

4

)10)(2(

a

aa Example 29:

abc

bca

32

)(128 324

Solution 6

2

8

1

53

54

20

4

102a

a

a

a

aa

Solution 523634

44

cbaabc

cba

Example 30:

0

1527

647

14

42

zyx

zyx

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.

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


LHS 14

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

1. 5)2v(4

31

2. 43cc21c 3. 1n51n5

4. c57

c1123

5. 2y5131y34 6. a232142a4

II. Solving inequalities in one variable. Solve each inequality.

1. 4b25b2b3 2. 1

8

x7

8

1x7

3.

10

1x5

5

3x3

4. 3

x1

4

8x

5. w2

4

w

5

120


LHS 15

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 =

3

2 , (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 16

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 17

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 18

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.

73

1834

yx

yx

Solve each system of equations by using elimination.

7. 15y3x7

27y6x5

8.

2yx3

2

7y2

1x

3

1


LHS 19

Exercises

V. Graphing a system of linear inequalities.

1. 3y

y2x3

2.

4x2y

1x2

1y

3. 13

622

yx

yx 4.

5y

5x


LHS 20

Exercises VI. Factoring.

1. 15a – 25b + 20 2. 48a3b

2 + 72a

2b

3 3. 9a

2 - 100

4. x2 + 5x + 4 5. a

2 + 10ab + 24b

2 6. 49a

2 + 28ab + 4b

2

VII. Simplifying Radicals.

1. 98 2. 60 3. 553

4. 87 5. 3

75 6.

23

4

VIII. Simplifying Expressions with Exponents.

1. ab

ba

5

45 24

2. ab

ba

5

2 72

3.

y

xy3

21

2

2

4.

0

2

4

5.0

2

xy

ba 5.

23

2 2

3 zy

x

z

yx 6.

321

225

8

24

yxx

zyxz


LHS 21

IX. Absolute Value Equations and Inequalities.

1. | | 2. | | 3. 2| |

4. | | 5.

| | 6. | |

7. | | 8. | | 9. 3| |

10.

| | 11. | | 12. |

|


LHS 22

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 23


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 24


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

1. 5 (3a – 5b + 4) 2. 24a2b

2 (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.

16

xy 4. 1 5.

3

532

z

yx 6.

5

33

y

z

IX. Absolute Value Equations and Inequalities. p. 20

1. -8, 8 2. No solution 3. 3,

4.

5.

6.

7.

8. 9. 10. 11. All real numbers are solutions

12.

Summer Review Packet For Algebra 2 CP/Honors...Algebra 2 builds on topics studied from both Algebra...

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