algebra 1 final exam review – 5 days (2nd semester)

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Algebra 1 Final Exam Review – 5 days (2nd Semester)

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Page 1: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Algebra 1 Final Exam Review – 5 days(2nd Semester)

Page 2: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Day 1

Page 3: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Solve an Inequality

- 5 -5

w < 3All numbers less than 3 are solutions to this problem!

0 5 10 15-20 -15 -10 -5-25 20 25

w + 5 < 8

Page 4: Algebra 1 Final Exam Review – 5 days (2nd Semester)

More Examples

-8 -8

r -10All numbers greater than-10 (including -10) ≥

0 5 10 15-20 -15 -10 -5-25 20 25€

8 + r ≥ −2

Page 5: Algebra 1 Final Exam Review – 5 days (2nd Semester)

More Examples

2 2

x > -1

All numbers greater than -1 make this problem true!

0 5 10 15-20 -15 -10 -5-25 20 25

2x > −2

Page 6: Algebra 1 Final Exam Review – 5 days (2nd Semester)

More Examples

-8 -8

2h ≤ 16

2 2

h ≤ 8

All numbers less than 8 (including 8)

0 5 10 15-20 -15 -10 -5-25 20 25

2h + 8 ≤ 24

Page 7: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Solve the inequality on your own.

1. x + 3 > -4

2. 6d > 24

3. 2x - 8 < 14

4. -2c – 4 < 2

x > -7

d ≥ 4

x < 11

c ≥ -3

Page 8: Algebra 1 Final Exam Review – 5 days (2nd Semester)

1132 x 1132 x3 3

82 x 82 x2 2

4x 4x

3 3

2 2?

?

?

?

Page 9: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Any time you multiply or divide both sides of an

inequality by a NEGATIVE, you must REVERSE THE

SIGN!!!!

TRY SOLVING THIS:

2695 x

Page 10: Algebra 1 Final Exam Review – 5 days (2nd Semester)

The solution would look likethis:

2695 x9 9

355 x5 5

7x

?

?

Page 11: Algebra 1 Final Exam Review – 5 days (2nd Semester)

SOLVE THIS:

3172 x7 7

242 x2 2

12x

Page 12: Algebra 1 Final Exam Review – 5 days (2nd Semester)

SOLVE THIS:

10134

x

13 13

34

x 434

4

x

12x

Page 13: Algebra 1 Final Exam Review – 5 days (2nd Semester)

SOLVE THIS:

1 1

9010 x

9x

89110 x

10 10

Page 14: Algebra 1 Final Exam Review – 5 days (2nd Semester)

3131

3

x

141531 x

SOLVE THIS:

15 15

3x

Page 15: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Solve the inequality and graph the solutions.

y ≤ 4y + 18

y ≤ 4y + 18–y –y

0 ≤ 3y + 18

–18 – 18

–18 ≤ 3y

To collect the variable terms on one side, subtract y from both sides.

Since 18 is added to 3y, subtract 18 from both sides to undo the addition.

Since y is multiplied by 3, divide both sides by 3 to undo the multiplication.

y –6

Page 16: Algebra 1 Final Exam Review – 5 days (2nd Semester)

4m – 3 < 2m + 6To collect the variable terms on one

side, subtract 2m from both sides.–2m – 2m

2m – 3 < + 6 Since 3 is subtracted from 2m, add 3

to both sides to undo the subtraction.

+ 3 + 3

2m < 9

Since m is multiplied by 2, divide both sides by 2 to undo the multiplication.

Solve the inequality and graph the solutions.

Page 17: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Solve the inequality and graph the solutions. Check your answer.

4x ≥ 7x + 6

–7x –7x

–3x ≥ 6

x ≤ –2

To collect the variable terms on one side, subtract 7x from both sides.

Since x is multiplied by –3, divide both sides by –3 to undo the multiplication. Change ≥ to ≤.

–10 –8 –6 –4 –2 0 2 4 6 8 10

The solution set is {x:x ≤ –2}.

Page 18: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Solve the inequality and graph the solutions. Check your answer.

5t + 1 < –2t – 6

5t + 1 < –2t – 6 +2t +2t

7t + 1 < –6

– 1 < –1

7t < –7

7t < –7

7 7

t < –1

–5 –4 –3 –2 –1 0 1 2 3 4 5

To collect the variable terms on one side, add 2t to both sides.

Since 1 is added to 7t, subtract 1 from both sides to undo the addition.

Since t is multiplied by 7, divide both sides by 7 to undo the multiplication.

The solution set is {t:t < –1}.

Page 19: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Solve the inequality and graph the solutions.

2(k – 3) > 6 + 3k – 3

2(k – 3) > 3 + 3k Distribute 2 on the left side of the

inequality.

2k – 6 > 3 + 3k

–2k – 2k

–6 > 3 + k

To collect the variable terms, subtract 2k from both sides.

–3 –3

–9 > k

Since 3 is added to k, subtract 3 from both sides to undo the addition.

k < -9

Page 20: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Day 2

Page 21: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Solving Compound InequalitiesSolving Compound Inequalities

752413 xorx

1

33

x

x6

122

x

x

61 xorx

Page 22: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Multiple Choice Solve:866 x

812.)

1214.)

1412.)

212.)

xd

xc

xb

xa+6 +6 +6

Page 23: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Multiple Choice Solve

7521783 xorx

13.)

13.)

63

25.)

13.)

xorxd

xorxc

xorxb

xorxa

−8 − 8

3x

3>

9

3x > 3

−5 − 5

2x

2≤

2

2x ≤1

Page 24: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Solving Compound InequalitiesSolving Compound Inequalities

5213 x

+1 +1 +1

−2

−2<

−2x

−2≤

6

−2

1 > x ≥ −3

−3 ≤ x <1

At the end you must flip the whole inequalityto have all the signs point to the left and lower numbers on the left

Page 25: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Solve

92513 x

−5 − 5 − 5

−18

−2≤

−2x

−2<

4

−2

9 ≥ x > −2

−2 < x ≤ 9

At the end you must flip the whole inequalityto have all the signs point to the left and lower numbers on the left

Page 26: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Solve for 3x + 2 < 14 and 2x – 5 > –11

Solve for 3x + 2 < 14 and 2x – 5 > –11 3x + 2 < 14 2x – 5 > -11 -2 -2 +5 +5 3x < 12 2x > -6

x < 4 AND x > -3

Page 27: Algebra 1 Final Exam Review – 5 days (2nd Semester)

• Solve

“Tree it up”: 5w + 3 = 7 OR 5w + 3 = -7

Solve both equations for w

5w + 3 = 7 5w + 3 = -7

5w = 4 5w = -10

w = w = -2

5 3 7w

45

Page 28: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Solve | x 2 | 5

x 2 IS POSITIVE

| x 2 | 5

x 7 x 3

x 2 IS NEGATIVE

| x 2 | 5

| 7 2 | | 5 | 5 | 3 2 | | 5 | 5

The expression x 2 can be equal to 5 or 5.

x 2 5

x 2 IS POSITIVE

x 2 5

Solve | x 2 | 5

The expression x 2 can be equal to 5 or 5. SOLUTION

x 2 5

x 2 IS POSITIVE

| x 2 | 5

x 2 5

x 7

x 2 IS POSITIVE

| x 2 | 5

x 2 5

x 7

x 2 IS NEGATIVE

x 2 5

x 3

x 2 IS NEGATIVE

| x 2 | 5

x 2 5

The equation has two solutions: 7 and –3.

CHECK

Page 29: Algebra 1 Final Exam Review – 5 days (2nd Semester)

2x 16

x = 8 x 1

| 2x 7 | -9

2x 7 IS POSITIVE

2x 7 9 2x 7 9

2x 7 IS NEGATIVE

2x 2

The equation has two solutions: 8 and –1.

Solve | 2x 7 | 5 4

Isolate the absolute value expression on one side of the equation.

SOLUTION

5 + 5

2x − 7 = 9

Page 30: Algebra 1 Final Exam Review – 5 days (2nd Semester)

• Solve 2 5 7x Subtract 5 from both sides

2 2x

2x = 2 2x = -2x = 1 x = -1

The solutions are -1 and 1

“TREE IT UP”

-5 -5

Page 31: Algebra 1 Final Exam Review – 5 days (2nd Semester)

This can be written as 1 x 7.

Solve | x 4 | < 3

x 4 IS POSITIVE x 4 IS NEGATIVE

x 4 3

x 7

x 4 3

x 1

Reverse inequality symbol!!!

The solution is all real numbers greater than 1 and less than 7.

“greatOR”“less thAND”

Page 32: Algebra 1 Final Exam Review – 5 days (2nd Semester)

2x 1 9

2x 10

2x + 1 IS NEGATIVE

x 5

Solve | 2x 1 | 3 6

2x 1 9

2x 8

2x + 1 IS POSITIVE

x 4

+3 +3

2x +1 ≥ 9

x ≥ 4 OR x ≤ -5

Page 33: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Examples

752 w

752 w 752 wor

122 w6w

22 w1wor

Check and verify on a number line. Numbers above 6 or below -1 keep the absolute value greater than 7. Numbers between them make the absolute

value less than 7.

Page 34: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Solve absolute-value inequalities.

Solve |x – 4| 5.

x – 4 is positive

x – 4 5

x – 4 is negative

x 9

Case 1: Case 2:

x – 4 –5

x –1

solution: –1 x 9

Page 35: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Solve absolute-value inequalities.

Solve |4x – 2| -18. Exception alert!!!!

When the absolute value equals a negative value, there is no solution.

Page 36: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Solve absolute-value inequalities.

Solve |2x – 6| 18.

2x – 6 is positive2x – 6 18

2x – 6 is negative

x 12

Case 1: Case 2:

2x - 6 –18

x –6Solution: –6 x 12

2x 24 2x –12

TRY THIS

Page 37: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Solve absolute-value inequalities.

Solve |3x – 2| -4. Exception alert!!!!

When the absolute value equals a negative value, there is no solution.

TRY THIS

Page 38: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Day 3

Page 39: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Solve: by ELIMINATIONx + y = 12

-x + 3y = -8We need to eliminate (get rid of) a variable.

The x’s will be the easiest. So, we will add the two equations.

4y = 4 Divide by 4

y = 1

Like variables must be lined under each other.

x + (1) =12

−1 −1

x =11

Then plug in the one value to find the other: ANSWER:

(11, 1)

Page 40: Algebra 1 Final Exam Review – 5 days (2nd Semester)

x + y =12

11 + 1 = 12

12 = 12

-x + 3y = -8

-11 + 3(1) = -8

-11 + 3 = -8

-8 = -8

Page 41: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Solve: by ELIMINATION5x - 4y = -21-2x + 4y = 18

We need to eliminate (get rid of) a variable.

The y’s be will the easiest.So, we will add the two equations.

3x = -3 Divide by 3

x = -1

Like variables must be lined under each other.

Then plug in the one value to find the other:

5(−1) − 4y = −21

−5 − 4 y = −21

+5 + 5

−4y

−4=

−16

−4y = 4

ANSWER:(-1, 4)

Page 42: Algebra 1 Final Exam Review – 5 days (2nd Semester)

5x - 4y = -215(-1) – 4(4) = -21

-5 - 16 = -21-21 = -21

-2x + 4y = 18

-2(-1) + 4(4) = 18

2 + 16 = 18

18 = 18

Page 43: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Solve: by ELIMINATION2x + 7y = 315x - 7y = - 45

We need to eliminate (get rid of) a variable.

The y’s will be the easiest. So, we will add the two equations.

7x = -14 Divide by 7

x = -2

Like variables must be lined under each other.

2(−2) + 7y = 31

−4 + 7y = 31

+4 + 4

7y

7=

35

7y = 5

Then plug in the one value to find the other: ANSWER:

(-2, 5)

Page 44: Algebra 1 Final Exam Review – 5 days (2nd Semester)

2x + 7y = 312(-2) + 7(5) = 31

-4 + 35 = 3131 = 31

5x – 7y = - 45

5(-2) - 7(5) = - 45

-10 - 35 = - 45

- 45 =- 45

Page 45: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Solve: by ELIMINATION

x + y = 30 x + 7y = 6

We need to eliminate (get rid of) a variable.

To simply add this time will not eliminate a variable. If one of the x’s was negative, it would be eliminated when we add. So we will multiply one equation by a – 1.

Like variables must be lined under each other.

Page 46: Algebra 1 Final Exam Review – 5 days (2nd Semester)

x + y = 30

x + 7y = 6( ) -1

x + y = 30

-x – 7y = -6

Now add the two equations and solve. -6y = 24

-6 -6

y = - 4

Then plug in the one value to find the other:

x + (−4) = 30

+ 4 + 4

x = 34

ANSWER:(34, -4)

Page 47: Algebra 1 Final Exam Review – 5 days (2nd Semester)

x + y = 3034 + - 4 = 30

30 = 30

x + 7y = 6

34 + 7(- 4) = 6

34 - 28 = 6

6 = 6

Page 48: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Solve: by ELIMINATION

x + y = 4 2x + 3y = 9We need to eliminate (get

rid of) a variable.

To simply add this time will not eliminate a variable. If there was a –2x in the 1st equation, the x’s would be eliminated when we add. So we will multiply the 1st equation by a – 2.

Like variables must be lined under each other.

Page 49: Algebra 1 Final Exam Review – 5 days (2nd Semester)

x + y = 4

2x + 3y = 9

-2x – 2y = - 8

2x + 3y = 9

Now add the two equations and solve. y = 1

( ) -2

Then plug in the one value to find the other:

x + (1) = 4

−1 −1

x = 3 ANSWER:(3, 1)

Page 50: Algebra 1 Final Exam Review – 5 days (2nd Semester)

x + y = 43 + 1 = 4 4 = 4

2x + 3y = 9

2(3) + 3(1) = 9

6 + 3 = 9

9 = 9

Page 51: Algebra 1 Final Exam Review – 5 days (2nd Semester)

1. Evaluate the following exponential expressions:

A. 42 =

B. 34 =

C. 23 =

D. (-1) =7

4 4 = 16

3 3 3 3 = 812 2 2 = 8

-1 -1 -1 -1 -1 -1 -1 = -1

REMEMBER TO PUT PARENTHESES AROUND NEGATIVE NUMBERS WHEN USING YOUR CALCULATOR!!!!!

Page 52: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Laws of Exponents

1. xm ⋅ x n = xm +n 2. xy( )m

= xm ym

3. xm( )

n= xmn 4.

x

y

⎝ ⎜

⎠ ⎟

m

=xm

ym

5.xm

x n = xm −n

Page 53: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Zero Exponents

• A nonzero based raise to a zero exponent is equal to one

a0 = 1

Page 54: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Negative Exponents

a-n= (

1______

an )

A nonzero base raised to a negative exponent is the reciprocal of the base raised to the positive exponent.

A nonzero base raised to a negative exponent is the reciprocal of the base raised to the positive exponent.

Page 55: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Basic Examples

32 xx 32x 5x

x 4( )

3= 34x 12x

Page 56: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Basic Examples

xy( )3

= 33 yx

3

y

x3

3

y

x

Page 57: Algebra 1 Final Exam Review – 5 days (2nd Semester)

4

7

x

x

1

47x 3x

7

5

x

x 57

1

x 2

1

x

Basic Examples

Page 58: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Examples

5

p

u

74 xx €

y 7( )

5=

3

9

x

x

1. 2.

3. 4.

u5

p5

y 7⋅5 =

y 35

x 4 +7 =

x11

x 9−3

1=

x11

Page 59: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Example

• Write 7,200,000 in scientific notation

Big Number means Positive Exponent

7.2 106

Page 60: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Example

• Write 476 in scientific notation.

Big Number means Positive Exponent

4.76 102

Page 61: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Example

Write 0.0062 in scientific notation.

Small Number means Negative Exponent

6.2 10-3

Page 62: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Example

1. Write these numbers in standard notation:

a.) 4.6 x 10ˉ³

b.) 4.6 x 10

2. Saturn is about 875,000,000 miles from the sun. What is this distance in scientific notation?

6

4.60 0

0.0046

4.6 0 0 0 0 0

4,600,000

8.75 108

Page 63: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Day 4

Page 64: Algebra 1 Final Exam Review – 5 days (2nd Semester)

4

16

25

100

144

= ±2

= ±4

= ±5

= ±10

= ±12

Page 65: Algebra 1 Final Exam Review – 5 days (2nd Semester)

8

20

32

75

40

=

= =

=

=

4 • 2

4 • 5

16 • 2

25 • 3

4 • 10

=

=

=

=

=

22

52

24

35

102

Perfect Square Factor * Other Factor

LE

AV

E I

N R

AD

ICA

L F

OR

M

Page 66: Algebra 1 Final Exam Review – 5 days (2nd Semester)

48

80

50

125

450

=

= =

=

=

16 • 3

16 • 5

25 • 2

25 • 5

225 • 2

=

=

=

=

=

34

54

5 2

55

215

Perfect Square Factor * Other Factor

LE

AV

E I

N R

AD

ICA

L F

OR

M

Page 67: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Solve x 2 = 8 algebraically. 1

2

12

x 2 = 8

SOLUTION

Write original equation.

x 2 = 16 Multiply each side by 2.

Find the square root of each side.x = 4

2 2

Page 68: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Solve using square roots. Check your answer.

x2 = 169

x = ± 13

The solutions are 13 and –13.

Solve for x by taking the square root of both sides. Use ± to show both square roots.

Substitute 13 and –13 into the original equation.

x2 = 169 (–13)2 169 169 169

Check x2 = 169 (13)2 169 169 169

x 2 = 169

Page 69: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Solve using square roots.

x2 = –49There is no real number whose

square is negative.

Answer: There is no real solution.€

x 2 = −49

Page 70: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Solve using square roots. Check your answer.

x2 = 121

x = ± 11

The solutions are 11 and –11.

Solve for x by taking the square root of both sides. Use ± to show both square roots.

Substitute 11 and –11 into the original equation.

x2 = 121 (–11)2 121

121 121

Check x2 = 121 (11)2 121 121 121

x 2 = 121

Page 71: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Solve using square roots. Check your answer.

x2 = 0

x = 0

The solution is 0.

Solve for x by taking the square root of both sides. Use ± to show both square roots.

Substitute 0 into the original equation.

Check x2 = 0 (0)2 0 0 0

x 2 = 0

Page 72: Algebra 1 Final Exam Review – 5 days (2nd Semester)

x2 = –16

There is no real number whose square is negative.

There is no real solution.

Solve using square roots. Check your answer.

x 2 = −16

Page 73: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Solve using square roots.

x2 + 7 = 7

–7 –7 x2 + 7 = 7

x2 = 0

The solution is 0.

Subtract 7 from both sides.

Take the square root of both sides.

x 2 = 0

Page 74: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Solve using square roots.

16x2 – 49 = 0

16x2 – 49 = 0+49 +49

Add 49 to both sides.

Divide by 16 on both sides.

Take the square root of both sides. Use ± to show both square roots.

Page 75: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Solve using the quadratic formula.0273 2 xx

a

acbbx

2

42

2 ,7 ,3 cba

)3(2

)2)(3(4)7()7( 2 x

6

24497 x

6

57x

6

12x

6

2x

3

1 ,2x

6

257x

x =7 + 5

6

x =7 − 5

6

Page 76: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Solve using the quadratic formulamm 1083 2

a

acbbm

2

42

8 ,10 ,3 cba

)3(2

)8)(3(4)10()10( 2 m

6

9610010 m

6

1410m

6

24m

6

4m

3

2 ,4 m

08103 2 mm

6

19610m

Page 77: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Solve using the quadratic formulaxx 16642

a

acbbx

2

42

64 ,16 ,1 cba

)1(2

)64)(1(41616 2 x

2

25625616 x

2

016x

8x

064162 xx

Page 78: Algebra 1 Final Exam Review – 5 days (2nd Semester)

1. Add the following polynomials:(9y - 7x + 15a) + (-3y + 8x - 8a)

Group your like terms.

9y - 3y - 7x + 8x + 15a - 8a

6y + x + 7a

Page 79: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Combine your like terms.

3a2 + 3ab + 4ab - b2 + 6b2

3a2 + 7ab + 5b2

2. Add the following polynomials:(3a2 + 3ab - b2) + (4ab + 6b2)

Page 80: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Just combine like terms. x2 - 3xy + 5y2

3. Add the following polynomials using column form:

(4x2 - 2xy + 3y2) + (-3x2 - xy + 2y2)

Page 81: Algebra 1 Final Exam Review – 5 days (2nd Semester)

You need to distribute the negative!!

(9y - 7x + 15a) + (+ 3y - 8x + 8a)

Group the like terms.

9y + 3y - 7x - 8x + 15a + 8a

12y - 15x + 23a

4. Subtract the following polynomials:

(9y - 7x + 15a) - (-3y + 8x - 8a)

Page 82: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Distribute the negative(7a - 10b) + (- 3a - 4b)

Group the like terms.7a - 3a - 10b - 4b

4a - 14b

5. Subtract the following polynomials:

(7a - 10b) - (3a + 4b)

Page 83: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Distribute the negative!!! 4x2 – 2xy + 3y2 + 3x2 + xy – 2y2

7x2 - xy + y2

6. Subtract the following:(4x2 - 2xy + 3y2) - (-3x2 – xy + 2y2)

Page 84: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Find the sum or difference.(5a – 3b) + (2a + 6b)

1. 3a – 9b

2. 3a + 3b

3. 7a + 3b

4. 7a – 3b

Page 85: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Find the sum or difference.(5a – 3b) – (2a + 6b)

1. 3a – 9b

2. 3a + 3b

3. 7a + 3b

4. 7a – 9b

Page 86: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Multiply (y + 4)(y – 3)

1. y2 + y – 12

2. y2 – y – 12

3. y2 + 7y – 12

4. y2 – 7y – 12

5. y2 + y + 12

6. y2 – y + 12

7. y2 + 7y + 12

8. y2 – 7y + 12

Page 87: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Multiply (2a – 3b)(2a + 4b)

1. 4a2 + 14ab – 12b2

2. 4a2 – 14ab – 12b2

3. 4a2 + 8ab – 6ba – 12b2

4. 4a2 + 2ab – 12b2

5. 4a2 – 2ab – 12b2

Page 88: Algebra 1 Final Exam Review – 5 days (2nd Semester)

5) Multiply (2x - 5)(x2 - 5x + 4)You cannot use FOIL because they are not BOTH binomials. You must use the

distributive property.

2x(x2 - 5x + 4) - 5(x2 - 5x + 4)

2x3 - 10x2 + 8x - 5x2 + 25x - 20

Group and combine like terms.

2x3 - 10x2 - 5x2 + 8x + 25x - 20

2x3 - 15x2 + 33x - 20

Page 89: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Multiply (2p + 1)(p2 – 3p + 4)

1. 2p3 + 2p3 + p + 4

2. y2 – y – 12

3. y2 + 7y – 12

4. y2 – 7y – 12

Page 90: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Multiply each:Multiply each:1) x+5( ) 2x−1( )

= 2x2 + 9x + -5

2) 3w−2( ) 2w−5( )

= 6w2 + -19w + 10

Page 91: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Multiply each:Multiply each:

3) 2a2 +a−1( ) 2a2 +1( )

4a4 + 2a3 + a - 1

Distribute the binomial

4a4 + 2a3 – 2a2 + 2a2 + a – 1

Page 92: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Use the FOIL method to multiply these binomials:

Multiply each:Multiply each:

1) (3a + 4)(2a + 1)

2) (x + 4)(x – 5)

3) (x + 5)(x – 5)

4) (c - 3)(2c - 5)

5) (2w + 3)(2w – 3)

= 6a2 + 3a + 8a + 4 = 6a2 + 11a + 4

= x2 - 5x + 4x - 20 = x2 - 1x - 20

= x2 - 5x + 5x - 25 = x2 - 25

= 2c2 - 5c - 6c + 15 = 2c2 - 11c + 15

= 4w2 - 6w + 6w - 9 = 4w2 - 9

Page 93: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Day 5

Page 94: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Review: What is the GCF of 25a2 and 15a?

5a

Let’s go one step further…

1) FACTOR 25a2 + 15a.

Find the GCF and divide each term

25a2 + 15a = 5a( ___ + ___ )

Check your answer by distributing.

225

5

a

a

15

5

a

a

5a 3

Page 95: Algebra 1 Final Exam Review – 5 days (2nd Semester)

2) Factor 18x2 - 12x3.

Find the GCF

6x2

Divide each term by the GCF

18x2 - 12x3 = 6x2( ___ - ___ )

Check your answer by distributing.

2

2

18

6

x

x

3

2

12

6

x

x

3 2x

Page 96: Algebra 1 Final Exam Review – 5 days (2nd Semester)

3) Factor 28a2b + 56abc2.

GCF = 28abDivide each term by the GCF

28a2b + 56abc2 = 28ab ( ___ + ___ )

Check your answer by distributing.28ab(a + 2c2)

228

28

a b

ab

256

28

abc

ab

a 2c2

Page 97: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Factor 20x2 - 24xy

1. x(20 – 24y)

2. 2x(10x – 12y)

3. 4(5x2 – 6xy)

4. 4x(5x – 6y)

Page 98: Algebra 1 Final Exam Review – 5 days (2nd Semester)

5) Factor 28a2 + 21b - 35b2c2

GCF = 7Divide each term by the GCF

28a2 + 21b - 35b2c2 = 7 ( ___ + ___ - _______ )

Check your answer by distributing.7(4a2 + 3b – 5b2c2)

228

7

a 21

7

b

4a2 5b2c2

2 235

7

b c

3b

Page 99: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Factor 16xy2 - 24y2z + 40y2

1. 2y2(8x – 12z + 20)

2. 4y2(4x – 6z + 10)

3. 8y2(2x - 3z + 5)

4. 8xy2z(2 – 3 + 5)

Page 100: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Factor each trinomial, if possible. The first four do NOT have leading coefficients, the last two DO have leading coefficients. Watch out for signs!!

1) t2 – 4t – 21 2) x2 + 12x + 32

1) x2 –10x + 24 4) x2 + 3x – 18

5) 2x2 + x – 21 6) 3x2 + 11x + 10

Factor These Trinomials!

(t – 7)(t + 3) (x + 8)(x + 4)

(x – 6)(x - 4) (x + 6)(x - 3)

2x2 + 7x – 6x – 21

x(2x + 7) – 3(2x + 7)

(x –3)(2x + 7)

3x2 + 6x + 5x + 10

3x(x + 2) + 5(x + 2)

(x + 2)(3x + 2)

REMEMBER YOU CAN CHECK YOUR ANSWERBY FOILING BACK OUT!!!!!

1 • -21 -1 • 213 • -7 -3 • 7

1 • 322 • 164 • 8

1 • 242 • 123 • 84 • 6

-1 • -24-2 • -12-3 • -8-4 • -6

1 • -18 -1 • 18 2 • -9 -2 • 93 • -6 -3 • 6

1 • -42, -1 • 42 2 • -21, -2 • 213 • -14, -3 • 146 • -7, -6 • 7

1 • 302 • 153 • 105 • 6

Page 101: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Solve this proportion.

____ = ____54

45x

Now cross

multiply

5x = 1805 5

x = 36

Page 102: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Solve the proportion.

____ = ____32

x18

Now cross

multiply

2x = 542 2

x = 27

Page 103: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Just solve…..

3

6 48

x10)

6x = 1446 6

x = 24

Page 104: Algebra 1 Final Exam Review – 5 days (2nd Semester)

11)4

2 16

m

16m = 8

16 16

m =8

16

NowReduce

1

2m

Page 105: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Use cross multiplying to solve the proportion.

1. 2520

= 45t

2. x9

= 1957

3. 23

= r36

4. n10

=288

t = 36

x = 3

r = 24

n = 35

Page 106: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Simplify:

9

2 4

2

2

x yz

xyz 3 3

3 8

x x y z

x y z z

3

8

x

z

Factor the numerator and denominator

Divide out the common factors.

Write in simplified form.

Page 107: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Simplify:

a

a a

3

4 32 a

a a

3

3 1( )( )

Factor the numerator and denominator

1

1a Divide out the common factors.

Write in simplified form.

Page 108: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Simplify

3 1 5

7 1 02

x

x x

3 5

5 2

( )

( )( )

x

x x

3

2x

Factor the numerator and denominator

Divide out the common factors.

Write in simplified form.

Page 109: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Simplify:

x x

x x

2

2

2 1 5

1 2

( )( )

( )( )

x x

x x

5 3

4 3

x

x

5

4

Factor the numerator and denominator

Divide out the common factors.

Write in simplified form.

Page 110: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Simplify:

1 4 3 5 2 1

1 2 3 0 1 8

2

2

x x

x x

7 2 5 3

6 2 6 3

2

2

( )

( )

x x

x x

7

6

Factor the numerator and denominator

Divide out the common factors.

Write in simplified form.

Page 111: Algebra 1 Final Exam Review – 5 days (2nd Semester)

Simplify each:

14 2

1 4

3 2

2 3.

y z

y z

23 6

2 1 2

2

. x

x

34 1 2

2 8

2

2.

c c

c c

=2 • 3 • 7 • y • y • y • z • z

2 • 7 • y • y • z • z • z

=3y

z

=(x + 6)(x − 6)

2(x − 6)

=x + 6

2

=(c + 6)(c − 2)

(c + 4)(c − 2)

=c + 6

c + 4