Dear Stuyvesant students,

This booklet was prepared by the Stuyvesant High School Mathematics

Department and contains prerequisite material for your math class. The materials

must be completed and brought to your math class on Wednesday, September 9

(the first day of school).

Directions:

Print and staple the pages.

Use the space provided to carry out your work.

Show all work.

Be neat. You may use pen or pencil.

A test based on the material will be given in your math class on Thursday,

September 17. No calculator will be permitted. It will be the first math test of the

term, and the test grade will be determined as follows:

Completion of the packet on time will account for 25% of the test grade.

Performance on the test will account for 75% of the test grade.

Don’t procrastinate. Solve a few problems each day. If you need extra help, please

go to www.khanacademy.org.

We are looking forward to seeing you this September!

Sincerely,

The Stuyvesant High School Mathematics Teachers

Stuyvesant High School Mathematics

Summer Homework

Due: September 9, 2015

Name ___________________________________

1. Solving Literal Equations

Model Problems

1. Solve for x: y = 17 - 2

x3

a) Multiply both sides by 3. 3y = 51 – 2x

b) Isolate 2x. 2x = 51 – 3y

c) Divide both sides by 2. x = 51 3y

2

or

x = 51 3

y2 2

2. Solve for c: b = a(c2 + d)

a) Expand the right side. b = ac2 + ad

b) Isolate ac2. ac2 = b – ad

c) Divide both sides by a. c2 = b ad

a

d) Take square roots. c = b ad

a

Solve for the variable indicated.

1) w 10e

D5

, for w

1) ____________________________

2) M = L ( 1 + bt ), for t 2)__________________________________

3) S = n

(a L)2

, for L

3) _____________________________

4) A = 2r , for r if r > 0

4) __________________________________

5) x2 + y2 = 9, for y 5) _______________________________

6) ( x + y ) 2 = z, for y 6) __________________________________

7) y = x 6

x

, for x

7) _______________________________

8) y = 2x 3

5 4x

, for x

8) __________________________________

2. Evaluating and Simplifying Expressions Containing Exponents

A. Evaluate each expression for n = 3 .

1) n3

1) ________

2) n2 – n3 2) _______

3) 4n2

3) ______

4) (4n)2 4)_______

5) –(n + 2)3

5)__________

6) (-n – 2)8 6) _______

7) (2 + n)15 7) _______

8) 3

n

4

8) _________

B. Write < or > in the blank if x > 0.

1) 322 _____ 325

2) 3x2 ____ (3x)2

3) (-5x)3 ___ 5x3

4) ( -2)16 ___ (-2)19 5)

103

4

_____ 7

3

4

C. Simplify each expression.

1) 2 3 2

3 2 3

( 36m n )

(12m n )

2) 7 2

3

( 9m )(8m )

6m

D. Evaluate.

1) 5-1 3-2

5) 3

1

2

3

2) (3-1 – 3-2)-1

6) (0.5)-3 + (0.5)o

3) 1 2

1 1

4 3

7) 1 o

4

4 4

4) 3 2

1 1

2 2

8) 1

2 3

2

2 2

3. Simplifying Expressions Containing Radicals

Model Problems

1. Simplify 2 48 .

2 48 =

2 16 3 =

2(4) 3 =

8 3 Answer

2. Simplify 8 2 50 .

8 2 50 =

2 2 2(5) 2 =

2 2 10 2 =

12 2 Answer

3. Simplify 6 2 .

6 2 =

12 =

2 3 Answer

4. Simplify 98

2.

98

2=

49 =

7 Answer

Simplify.

1) (2 24)(5 6)

6) 6 50

5 18

2) ( 3 20)( 2 5)

7) 3 8

3) 5 28 3 63 2 112

8) 5 32

4) 8 24 2 54 28

9) 4 81

5) 3 20

2 10

10) 7

19

4. Polynomial Operations

Model Problems

1. Simplify: 15y2 – 3y – (6y3 – 2y2 + 7y)

15y2 – 3y – 6y3 + 2y2 – 7y =

-6y3 + 17y2 - 10y Answer

2. Find the product: (x – 3)(x + 7)

x2 + 7x - 3x – 21 =

x2 + 4x – 21 Answer

3. Expand: (2x + 5)2

(2x + 5)(2x + 5) =

4x2 + 10x + 10x + 25 =

4x2 +20x + 25 Answer

4. Find the product: (5 – 3y)(4 + 7y)

20 + 35y – 12 y – 21y2 =

-21y2 + 23y + 20 Answer

1) Simplify: (4x – 3) – (12x + 1) – (-3x + 4) 1) ______________________

5) Multiply: (2x – 3)(x2 – 6x + 4) 5) _______________________

2) Simplify: 2a – 5a4 +6a2 – 4a2 +7a + 8a4 – a3 2) ______________________

6) Multiply: (3x + 2)(3x – 2) 6) ______________________

3) Simplify: 2x2y2 – 3(x2 – 2) + y2(x2 – 3) + 7 3) _________________________

7) Multiply: (7 + 12x)(7 – 12x) 7) ________________________

4) Multiply: (5x + 3y)(4x – y) 4) __________________________

8) Expand: (10-3c)2

8) ____________________

5. Factoring Out Common Factors

Model Problems

Factor:

1) 3xy2 + 6x2y – 9xy

3xy ( y + 2x – 3) Answer

2) 24m3 + 6m2 – 12m

6m ( 4m2 + m – 2) Answer

3) 6t (3t – 1) + 9 (3t – 1)

(6t + 9) (3t – 1) =

3 (2t + 3) (3t – 1) Answer

4) a2 + 2a + ab + 2b

a (a + 2) + b (a + 2)=

(a + b) (a + 2) Answer

Factor:

1) 18a – 6ab2

1)_________________________________

2) -12p2q3 – 20p3q3 + 8p2q4

2) ___________________________________

3) a2 (a + 2) – a (a + 2) 3) __________________________________

4) 3x3 (x – 1) – 6x2 (x – 1)

4)___________________________

5) 3x – xy + 3y – y2

5) __________________________

6. Factoring Trinomials of the Form x2 + bx + c

Model Problems

Factor.

1. x2 + 5x + 6

Find two numbers whose sum is 5 and

whose product is 6.

The numbers are 2 and 3.

x2 + 5x + 6 = ( x + 2 ) ( x + 3 ) Answer

2. a2 – 8a + 12

___ × ___ = 12 and ___ + ___ = -8

The numbers are -6 and -2.

a2 – 8a + 12 = ( a – 6) (a – 2) Answer

3. m2 – 5mn – 14n2

___ × ___ = -14 and ___ + ___ = -5

The numbers are -7 and 2

m2 – 5mn – 14n2 = (m – 7n) (m + 2n) Answer

4. n2 – 14n + 49

___ × ___ = 49 and ___ + ___ = -14

The numbers are -7 and -7

n2 -14n + 49 = (n – 7 ) ( n – 7)

= (n – 7)2 Answer

5. 3x2 – 18x + 24

First factor out the greatest common factor.

3( x2 – 6x + 8 )=

3 ( x – 4) ( x – 2) Answer

Factor completely.

1. x2 + 16x + 63

6. 6 – x – x2

2. x2 – 18x + 32

7. 2x2 + 12x + 10

3. k2 – 4k – 21

8. 4a2 – 16a - 20

4. x2 + 7x – 60

9. x3 – 2x2 – 3x

5. 56 + t – t2

10. 7a2 – 35a + 42

7. Factoring Trinomials of the Form ax2 + bx + c with a > 1

Model Problems

Factor:

1. 2y2 – 11y + 12

Multiply a and c: (2)(12) = 24

Find two numbers whose sum is -11 and whose

product is 24.

The numbers are -8 and -3.

2y2 – 11y + 12 =

2y2 – 8y – 3y +12 =

2y(y – 4) – 3(y – 4) =

(2y – 3)(y – 4) Answer

2. 4a2 + 12a + 9

___ x ___ = 36 and ___ + ___ = 12

The numbers are 6 and 6.

4a2 +12a + 9 =

4a2 + 6a + 6a + 9 =

2a (2a + 3) + 3(2a + 3) =

(2a + 3)(2a + 3) =

(2a + 3)2 Answer

3. 5a2 + a – 4

___ x ___ = -20 and ___ + ___ = 1

The numbers are 5 and -4.

5a2 +a – 4 =

5a2 + 5a – 4a – 4 =

5a(a + 1) – 4(a + 1) =

(5a – 4)(a + 1) Answer

4. 20x2 + 14x – 24 =

2(10x2 + 7x – 12) = (Factor out GCF first.)

2(5x – 4)(2x + 3) Answer

Factor completely. Remember to factor our greatest common factor first.

1. 2n2 + 7n + 6

5. 5u2 _ 39u – 8

2. 9k2 + 12k + 4

6. 10p2 + p – 21

3. 25a2 – 20a + 4

7. 35x2 – 66x + 16

4. 3g2 – 14g + 8

8. 18n2 + 30n – 12

8. Factoring Completely

Factor each polynomial completely.

1) 16 – 81x4 1) ____________________________

2) x2y – 16y 2)__________________________________

3) x3 + 5x2 – 14x 3) _____________________________

4) x4 – 7x2 +12 4) __________________________________

5) 12x2 + 2x – 4 5) _______________________________

6) 4x3 -10x2 +6x 6) __________________________________

7) 12x3y – 5x2 y2 – 2xy3

7) _______________________________

8) 9 – 9x – 54x2

8) __________________________________

Model Problems

Factor completely.

1. 3x2 – 75

a) Factor out the greatest common factor. 3(x2 – 25)

b) Factor the difference of two squares. 3(x + 5)(x-5) Answer

2. 4x3 – 20x2 + 24x

a) Factor out the greatest common factor. 4x(x2 – 5x + 6)

b) Factor the trinomial. 4x(x – 2)(x – 3) Answer

3. x4 – 16

a) Factor the difference of two squares. (x2 + 4)(x2 – 4)

b) Factor the difference of two squares. (x2 + 4)(x +2)(x – 2) Answer

9. Solving Quadratic Equations by Factoring, or, if b = 0, by Taking Square Roots

Find the solution set.

1) x2 – x – 6 = 0

Factor the left-hand side: (x – 3)(x + 2) = 0

If AB = 0, then A = 0 or B = 0: x – 3 = 0 or x +

2 = 0

x = 3 or x = -

2

The solution set is {-2, 3}. Answer

2) 2x 2 = 6 – x

Write in ax2 + bx + c = 0 form: 2x2 + x – 6= 0

Factor the left-hand side: (2x – 3)(x + 2) =

0

2x – 3 = 0 or x +

2 = 0

x = 3/2 or x = -2

The solution set is {-2, 3/2}. Answer

3) 4x2 = 75 + x2

Since there is no x term (that is, if written in

standard ax2 + bx + c = 0 form, b = 0), we can

solve by taking square roots. First solve for x2

and then take square roots.

3x2 = 75

x2 = 25

x = ±5

The solution set is {-5, 5}. Answer.

Alternately, we may solve by factoring.

3x2 – 75 = 0

3 ( x2 – 25 ) = 0

3(x + 5)(x – 5) = 0

x + 5 = 0 or x – 5 = 0

x = -5 or x = 5

Find the solution set (continued on next page).

1) x2 – 9x + 25 = 5

2) x2 – 8x – 40 = 4 – x

3) 3x2 + 6x – 72 = 0

4) 3x2 = 2 - x

5) 4x2 = 3x + 1

6) 2x2 – 128 = 0

7) 6x2 = 2x [Hint: Do not divide both sides by x.]

8) 4x (x + 3) = -9

10. Solving a System of Linear Equations by Addition or Subtraction (Elimination Method)

Model Problems

Solve:

1. 3x – 5y = -9

4x + 5y = 23

We may eliminate y by adding the equations to

get

7x = 14

x = 2

Now, substitute 2 for x in one of the given

equations:

4(2) + 5y = 23

8 + 5y = 23

5y = 15

y = 3

The solution of this linear system is the ordered

pair (2, 3). Answer

2. 2x + 3y = 8

-5x + 4y = 6

Adding or subtracting will not eliminate a

variable.

Multiply the first equation by 4 and the second

by 3:

8x + 12y = 32

-15x + 12y = 18

Subtract the resulting equations to get rid of y:

23x = 14

x = 14/23

Now, we may substitute 14/23 for x in the first

or second given equation to solve for y, or

instead solve for y by eliminating x. We may do

this by multiplying the first equation by 5 and

the second by 2 and then adding. We get y =

52/23.

The solution of this system is (14/23, 52/23).

Answer

Solve by the elimination method (continued on next page).

1) 8x – 3y = 38 4x – 5y = 26

2) 4x + 3y = 3 3x – 2y = -19

3) 1 1 1

x y3 2 2

1 1 27

x y5 3 5

4) 3 y 11

x4 3 2

2x 3y 21

5 2 10

11. Solving a System of Linear Equations by Substitution

Model Problem

Solve:

2x – y = 13

4x + 3y = 1

In the first equation, solve for y in terms of x:

y = 2x – 13

Substitute this expression for y in the second given equation and solve for x:

4x + 3 (2x – 13) = 1

4x + 6x – 39 = 1

10x – 39 = 1

10x = 40

x = 4

Substitute this value of x in the equation y = 2x – 13 to find y:

y = 2 (4) – 13

y = 8 -13

y = -5

The solution of the system is (4, -5). Answer

Solve by the substitution method (continued on next page).

1) x + 5y = -11 4x – 3y = 25

2) 3x – 4y = -15 5x + y = -2

3) 4x – 3y = 9 2x – y = 5

4) x y

12 3

x 2y

14 3

12. The Slope-Intercept Form of an Equation of a Line

Model Problems

1) Determine whether the graphs of the equations are parallel, perpendicular or neither.

a) y = 2x + 3

2y – 4x = 7

If y = mx + b, then the slope of the graph is m. In the second equation, solve for y in terms of x, and get

y = 2x + 7/2. Therefore, the graph of each equation has slope 2. If two lines have equal slopes, then they

are parallel.

Therefore the graphs of the equations are parallel lines. Answer

b) 3x – 5y = 10

5x + 3y + 8 = 0

Rewrite each equation in y = mx + b form.

y = (3/5) x – 2

y = (- 5/3) x – 8/3

Since the slopes of the lines are negative reciprocals (3/5 and -5/3), the lines are perpendicular. Answer

2) Write an equation of the line that contains the point (3, 4) and has x-intercept -3.

We need to find m and b so that we can write an equation of the form y = mx + b. (If you know the point-

slope form of a line, then you may use that instead.)Since the x-intercept is -3, the point (-3, 0) is a point on

the line.

m =

2 1

2 1

y y

x x=

0 4

3 3=

2

3

y = mx + b

y = 2

3x + b

Since (3, 4) is a point on the line, x = 3, y = 4 satisfies the equation of the line. (Or you may use the other

point,(-3, 0) instead).

4 = 2

3(3) + b

b = 2

So, an equation of the line is y 2

3x + 2.

Problems are on the next page.

1) State the slope and y-intercept of the graph of x + 2y – 5 = 0.

2) State the slope and y-intercept of the graph of 4x – 7y + 15 = 0.

3) Write an equation of a line that is parallel to the graph of 2x – 3y = 8 and has the same y-intercept as the graph of x + 6y – 9 =12.

4) Write an equation of the line that is perpendicular to the graph of 7x + 10y = 1 and passes through the origin.

5) Write an equation of the line that passes through the points (-1, 5) and (2, -1).

Extra Credit: The equations of the lines containing three sides of

a square are y = 1

x 82

, y = 1

x 22

, and

y = 2x 12 . Write an equation of the line

containing the fourth side. Use the back of the page if there is not enough room here for your solution.