Algebra 1 Final Exam Review – 5 days (2nd Semester)
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Transcript of Algebra 1 Final Exam Review – 5 days (2nd Semester)
Algebra 1 Final Exam Review – 5 days(2nd Semester)
Day 1
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
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
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
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
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
1132 x 1132 x3 3
82 x 82 x2 2
4x 4x
3 3
2 2?
?
?
?
Any time you multiply or divide both sides of an
inequality by a NEGATIVE, you must REVERSE THE
SIGN!!!!
TRY SOLVING THIS:
2695 x
The solution would look likethis:
2695 x9 9
355 x5 5
7x
?
?
SOLVE THIS:
3172 x7 7
242 x2 2
€
12x
SOLVE THIS:
10134
x
13 13
34
x 434
4
x
12x
SOLVE THIS:
1 1
9010 x
9x
89110 x
10 10
3131
3
x
141531 x
SOLVE THIS:
15 15
3x
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
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.
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}.
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}.
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
Day 2
Solving Compound InequalitiesSolving Compound Inequalities
752413 xorx
1
33
x
x6
122
x
x
61 xorx
Multiple Choice Solve:866 x
812.)
1214.)
1412.)
212.)
xd
xc
xb
xa+6 +6 +6
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
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
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
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
• 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
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
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
• 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
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”
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
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.
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
Solve absolute-value inequalities.
Solve |4x – 2| -18. Exception alert!!!!
When the absolute value equals a negative value, there is no solution.
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
Solve absolute-value inequalities.
Solve |3x – 2| -4. Exception alert!!!!
When the absolute value equals a negative value, there is no solution.
TRY THIS
Day 3
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)
x + y =12
11 + 1 = 12
12 = 12
-x + 3y = -8
-11 + 3(1) = -8
-11 + 3 = -8
-8 = -8
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)
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
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)
2x + 7y = 312(-2) + 7(5) = 31
-4 + 35 = 3131 = 31
5x – 7y = - 45
5(-2) - 7(5) = - 45
-10 - 35 = - 45
- 45 =- 45
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.
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)
x + y = 3034 + - 4 = 30
30 = 30
x + 7y = 6
34 + 7(- 4) = 6
34 - 28 = 6
6 = 6
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.
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)
x + y = 43 + 1 = 4 4 = 4
2x + 3y = 9
2(3) + 3(1) = 9
6 + 3 = 9
9 = 9
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!!!!!
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
Zero Exponents
• A nonzero based raise to a zero exponent is equal to one
a0 = 1
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.
Basic Examples
32 xx 32x 5x
€
x 4( )
3= 34x 12x
Basic Examples
€
xy( )3
= 33 yx
3
y
x3
3
y
x
4
7
x
x
1
47x 3x
7
5
x
x 57
1
x 2
1
x
Basic Examples
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
Example
• Write 7,200,000 in scientific notation
Big Number means Positive Exponent
7.2 106
Example
• Write 476 in scientific notation.
Big Number means Positive Exponent
4.76 102
Example
Write 0.0062 in scientific notation.
Small Number means Negative Exponent
6.2 10-3
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
Day 4
4
16
25
100
144
= ±2
= ±4
= ±5
= ±10
= ±12
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
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
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
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
Solve using square roots.
x2 = –49There is no real number whose
square is negative.
Answer: There is no real solution.€
x 2 = −49
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
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
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
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
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.
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
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
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
1. Add the following polynomials:(9y - 7x + 15a) + (-3y + 8x - 8a)
Group your like terms.
9y - 3y - 7x + 8x + 15a - 8a
6y + x + 7a
Combine your like terms.
3a2 + 3ab + 4ab - b2 + 6b2
3a2 + 7ab + 5b2
2. Add the following polynomials:(3a2 + 3ab - b2) + (4ab + 6b2)
Just combine like terms. x2 - 3xy + 5y2
3. Add the following polynomials using column form:
(4x2 - 2xy + 3y2) + (-3x2 - xy + 2y2)
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)
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)
Distribute the negative!!! 4x2 – 2xy + 3y2 + 3x2 + xy – 2y2
7x2 - xy + y2
6. Subtract the following:(4x2 - 2xy + 3y2) - (-3x2 – xy + 2y2)
Find the sum or difference.(5a – 3b) + (2a + 6b)
1. 3a – 9b
2. 3a + 3b
3. 7a + 3b
4. 7a – 3b
Find the sum or difference.(5a – 3b) – (2a + 6b)
1. 3a – 9b
2. 3a + 3b
3. 7a + 3b
4. 7a – 9b
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
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
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
Multiply (2p + 1)(p2 – 3p + 4)
1. 2p3 + 2p3 + p + 4
2. y2 – y – 12
3. y2 + 7y – 12
4. y2 – 7y – 12
Multiply each:Multiply each:1) x+5( ) 2x−1( )
= 2x2 + 9x + -5
2) 3w−2( ) 2w−5( )
= 6w2 + -19w + 10
Multiply each:Multiply each:
3) 2a2 +a−1( ) 2a2 +1( )
4a4 + 2a3 + a - 1
Distribute the binomial
4a4 + 2a3 – 2a2 + 2a2 + a – 1
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
Day 5
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
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
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
Factor 20x2 - 24xy
1. x(20 – 24y)
2. 2x(10x – 12y)
3. 4(5x2 – 6xy)
4. 4x(5x – 6y)
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
Factor 16xy2 - 24y2z + 40y2
1. 2y2(8x – 12z + 20)
2. 4y2(4x – 6z + 10)
3. 8y2(2x - 3z + 5)
4. 8xy2z(2 – 3 + 5)
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
Solve this proportion.
____ = ____54
45x
Now cross
multiply
5x = 1805 5
x = 36
Solve the proportion.
____ = ____32
x18
Now cross
multiply
2x = 542 2
x = 27
Just solve…..
3
6 48
x10)
6x = 1446 6
x = 24
11)4
2 16
m
16m = 8
16 16
m =8
16
NowReduce
1
2m
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
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.
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.
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.
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.
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.
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