```CST Prep Part II
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10
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25.1
Designed by Ms.Carranza and Mrs. Murray Solved by: 8th Grade Gate Students 2011
Standard 10.0
Students add, subtract, multiply, and divide monomials and polynomials.
Students solve multi-step problems, including word problems, by using
these techniques.
Problem 47
Problem 48
Problem 49
Problem 50
Problem 51
Problem 52
Rules &amp;
Strategies
Vocabulary
51) A volleyball court is shaped like a rectangle. It has a
width of x meters and length of 2x meters. Which
expression gives the area of the court in the square
meters?
A
B
C
D
Solution &amp;
3x
2x&sup2;
3x&sup2;
2x&sup3;
Standard 10
Vocabulary
• Expression : a mathematical phrase that
contains operations, numbers, and/or
variables.
• Area: The number of non overlapping unit
squares of a given size that will exactly cover
the interior of a plane figure.
• Square Meters: a unit of area
measurement equal to a square measuring
one meter on each side.
Back to
Problem
Rules &amp; Strategies
• Area= base (length) * height (width)
• Product of Powers: combine base, add
1
powers
• Add exponents ( x = x )
x
2x
Back to
Problem
Solution
2x • 1x
1
2x
1
2
Back to
Problem
Standard 10
Vocabulary
Solution
and
50) Which of the following
expressions
is equal to (x+2)+(x-2)(2x+1)?
Rules and
Strategies
Standard
10
(x+2)+(x-2x)(2x+1)?
Step 1: Area Method
2x + 1
x
-2
Step 2: Combine like
terms and put in
Descending order
2x&sup2;
+x
-4x
-2
2x&sup2;-3x-2
1X + 2 + 2x&sup2; - 3x - 2
0+
2x&sup2;
-2x
Back to
Problem
Standard 10
Vocabulary
Area Method:
(ax + b)(ax + b)
ax + b
ax
b
aax
abx
abx
bb
aax + (abx + abx) + bb
Descending Order: ordering terms from greatest to least
(ex. 1x+ 2x&sup2; + 3x+1x&sup3; + 9 = 1x&sup3; + 2x&sup2; + 3x + 1x + 9)
Back to
Problem
Rules &amp; Strategies
1st. Use the Area Method
2nd. Compare like terms
3rd. Descending Order
Back to
Problem
Rules &amp;
Strategies
Vocabulary
49)The sum of the two binomials is 5x2-6x, If
one of the binomials is 3x2-2x, what is the
other binomial
A 2x2-4x
B 2x2-8x
C 8x2+4x
D 8x2-8x
Solution and
Standard 10
Vocabulary
• Binomial- A polynomial with two terms
Back to
Problem
Rules and strategies
• Line up the binomials
• Combine both binomials by subtracting
Back to
Problem
Solution
1st binomial
2nd binomial
2
(3x -2x)
2
+2x -4x
2
5x -6x
Sum of binomials
Back to
Problem
Standard 10
Rules &amp;
Strategies
Vocabulary
3
5x
47)
7
10x
A) 2x4
B)1
2x4
C) 1
5x4
D) X4
5
Solution and
Standard 10.0
Vocabulary
• Quotient of Powers:
simplify coefficients
combine base
subtract exponents
Back to
Problem
Rules &amp; Strategies
• In order to skip the negative exponent rule,
circle the biggest exponent to tell if the
variable stays in the denominator or
numerator.
• Subtract powers
• Divide if there is any whole numbers in the
numerator and denominator.
Back to
Problem
Solution
5x3
1
10x7
2x4
Back to
Problem
Standard 10.0
Rules and
Strategies
Vocabulary
48.) (4x&sup2;-2x+8) – (x&sup2;+3x-2)
A) 3x&sup2;+x+6
B) 3x&sup2;+x+10
C) 3x&sup2;-5x+6
D) 3x&sup2;-5x+10
Solution and
Standard 10
Vocabulary
• Parenthesis ( ); indicates separate grouping of
symbols
• Exponent &sup2;; a symbol or number placed above and
after another symbol or number to denote the
power to which the latter is to be raised
• Variable ‘x’; a quantity or function that may assume
any given value or set of values
Back to
Problem
Rules and Strategies
• Subtracting Polynomials / Lesson 7-7
• Distribute (-) before grouping
Back to
Problem
(4x&sup2;-2x+8) – 1 (x&sup2;+3x-2)
(4x&sup2;-2x+8)-x&sup2;-3x+2
(4x&sup2;-1x&sup2;) + (-2x-3x) + (8+2)
3x&sup2;-5x+10
Back to
Problem
Standard 10
Standard 11
Students solve multistep problems involving linear equations and
inequalities in one variable.
Problem 53
Problem 54
Problem 55
Problem 56
Rules &amp;
Strategies
Vocabulary
53)Which is the factored form of
3a&sup2;-24ab+48b&sup2;?
a. (3a-8b)(a-6b)
b. (3a-16b)(a-3b)
c. 3(a-4b)(a-4b)
d. 3(a-8b)(a-8b)
Solution &amp;
Standard 11
Vocabulary
• Factored form= Form of equation in which
each term is simplified from factoring
methods and GCF.
Back to
Problem
Rules &amp; Strategies
• Ask yourself if you can factor out a GCF
• Use the diamond method
• Keep asking yourself whether you
can factor more.
If there is a GCF, don’t forget to
Back to
Problem
Factor out a gcf
3a&sup2;-24ab + 48b&sup2;
3 3
3
3(a&sup2;-8ab+16b&sup2;)
+16
keep factoring!!
a
a
-4b
-4b
3 (a-4b)2
Standard 11
Back to
Solution: C
-8
Problem
Rules &amp;
Strategies
Vocabulary
54) Which is the factor of x&sup2; – 11x + 24
A
B
C
D
Solution &amp;
x+3
x-3
x+4
x-4
Standard 11
Vocabulary
• Factor = A number or expression that is
multiplied by another number or expression
to get a product.
• Term = a part of an expression that is added or
subtracted
• Binomial = 2 terms
• Trinomial = 3 terms
Back to
Problem
Rules &amp; Strategies
•
•
•
•
•
•
•
•
•
•
Since there are three terms in this trinomial you use the “Diamond Method”
Diamond Method Formula : ax&sup2; + bx + c
Make an “X” on your paper.
On the top intersect write a multiplication sign (x) to show that the two
denominators multiply to equal the number you're going to get above the
multiplication sign.
On the bottom intersect write a plus sign (+) to show that the two
denominators added together equal the number you’re going to get below the
Plug the value for “c” above the multiplication sign ( +24 )
Plug in the value for “b” below the addition sign ( -11 )
Now , think of two numbers that multiplied equal +24 and added equal -11 .
( You should come up with -3 and -8 )
Lastly , you rewrite your fractions as binomials : 1x = ( x – 3 )
-3
Back to
Problem
Solution
x&sup2; – 11x + 24
+ 24
1x
x 1x
-3
+ -8
- 11
( x – 3 ) ( x – 8 )  Final answer
Back to
Problem
Standard 11
Rules &amp;
Strategies
Vocabulary
55) Which of the following shows
2
9t + 12t + 4 factored completely?
2
A (3t + 2)
B (3t + 4) (3t +1)
C (9t + 4) (t + 1)
2
D 9t + 12t + 4
Solution &amp;
Standard 11
Vocabulary
• Factored :one of two or more numbers,
algebraic expressions, or the like, that when
multiplied together produce a given product;
a divisor.
Back to
Problem
Rules &amp; Strategies
• Always ask yourself which factoring method
should I use?
• You should always see if you can factor out a
GCF
• Use“diamond method” when there is no GCF
and the coefficient is greater than one.
Back to
Problem
Solution
•Make sure to
simplify
the fractions
2
9t + 12t + 4
+36
3t
2
3
3
9t
+6
9t
+6
3
3
3t
2
+12
(3t+2)
2
Back to
Problem
Standard 11
Rules &amp;
Strategies
Vocabulary
56) What is the complete
factorization of 32-8z&sup2;
A. -8(2+z)(2-z)
B. 8(2+z)(2-z)
C. -8(2+z)&sup2;
D. 8(2-z)&sup2;
Solution &amp;
Standard 11
Vocabulary
• Factorization: to simplify
Back to
Problem
Rules &amp; Strategies
•
•
•
•
•
•
•
Find GCF
Divide each term by GCF
Rewrite equation w/ GCF outside of parenthesis
Divide inside terms by a negative
Rewrite the equation w/ negative on the outside
Difference of 2 Squares
Back to
Problem
Solution
32-8z&sup2;
-8 -8
factor out a -8 to get a positive z&sup2;
-8(z&sup2;-4) keep factoring diff of 2 squares
-8(z+2)(z-2)
Or
-8(2+z)(2-z)
Standard 11
Back to
Problem
Standard 12
• Students simplify fractions with polynomials in
the numerator and denominator by factoring
both and reducing them to the lowest terms.
Problem 77
Problem 81
Problem 78
Problem 79
Problem 80
Rules and
Strategies
Vocabulary
78.) 6x2 + 21x + 9
4x2 - 1
Solution and
A)3(x+1)
2x-1
B) 3(x+3)
2x-1
C) 3(2x+3)
4(x-1)
D) 3(x+3)
2x+1
Standard 12
Vocabulary
• GCF ; Greatest Common Factor
• Super Diamond ; x2 + bx + x
• Difference of Two Squares ; terms need to be in
perfect squares [Example: (a+b)(a-b)]
Back to
Problem
Rules and Strategies
• Top ; find GCF, super diamond
• Bottom ; difference of two squares, terms
need to be perfect squares
Back to
Problem
6x2 + 21x + 9
4x2 – 1
1) GCF
2) Difference of
Two Squares
3)Cross Cancel
Top/Numerator ;
6x2 + 21x + 9
Top/Numerator ;
6x2: 2 ∙ 3 ∙ x ∙ x
21x: 3 ∙ 7 ∙ x
x = +2
9: 3 ∙ 3
x
+3
GCF: 3
+6
+6
∙
+
+2
x+1
+7
Bottom/Denominator ;
4x2 - 1
2x 2x 1
1
(2x+1)(2x-1)
Cross Cancel
3(x+3)(2x+1) =
(2x+1)(2x-1)
6x 2 +21 + 9
3(x+3)(2x+1)
3
3 3
3 ( 2x 2 + 7x + 3)
Super Diamond!!
Back to
Problem
3(x+3)
2x-1
Standard 12
Rules &amp;
Strategies
Vocabulary
79)What is x2-4x+4 reduced to lowest terms?
x2-3x+2
Solution &amp;
A) x-2
X-1
B) x-2
X+1
C) x+2
X-1
D) x+2
X+1
Standard 12
Vocabulary
• Reduce = lower in degree
• Lowest term = the form of a fraction after
dividing the numerator and denominator by
their greatest common divisor.
Back to
Problem
Rules &amp; Strategies
Back to
Problem
• First you should ask yourself which strategy is
best to change the equation into a dividable state
using either; Greatest common facture (GCF),
difference of 2 squares, or one of the diamond
methods.
• In this case the best one would be the diamond
method on both the denominator and numerator.
• Once you find the factors of both of the
trinomials reduce by dividing the like terms.
Solution
ax2+bx+c=0
Step 1: x2-4x+4
x2-3x+2
Step 2: x2-4x+4 (x-2) (x-2)
Diamond Method
4
x
*
x
-2
-2
x2-3x+2
+
-4
Divide out!
(x-2) (x-2) = (x-2)
(x-2) (x-1) (x-1)
(x-2) (x-1)
2
x *
-2
x
-1
+
-3
Back to
Problem
Standard 12
Rules and
Strategies
Vocabulary
80. What is 12a3 – 20a2 reduced to lowest terms?
16a2 + 8a
Solution and
A) a
2
B) 3a – 5
2a + 1
C) -2a
4 + 2a
D) a (3a – 5)
2 (2a + 1)
Standard 12
Vocabulary
• Reduced ; simplified or lowest terms
Back to
Problem
Rules and Strategies
• Find GCF for both numerator and
denominator
• Simplify to lowest terms
Back to
Problem
12a3 - 20a
16a2 + 8a
Solution
and
2
2
4 a2 (3a – 5)
8 a (2a + 1)
a (3a – 5)
2 (2a +1)
GCF:4a
GCF: 8a2
Numerator :
12a3 - 20a2
4a2
4a2
4a2 (3a – 5)
*Simplify
Denominator:
16a2 + 8a
8a
8a
8a (2a + 1)
Back to
Problem
Standard 12
Rules &amp;
Strategies
Vocabulary
81) What is the simplest for of the
fraction:
_x2-1_
x2+x-2
Solution &amp;
A)_-1__
x-2
B)x-1
x-2
C)x-1
x+2
D)x+1
x+2
Standard 12
Vocabulary
• Simplest form of a rational expression- A
rational expression is in simplest form if the
numerator and denominator have no common
factors. Ex:
2
x -1
(x-1)(x+1)
2
x +x-2
(x+1)(x+2)
(x+1)
(x+2)
Simplest Form
Back to
Problem
Rules &amp; Strategies
1)How will I factor?
2)Simplify/ Divide out
1.GCF
2.Diamond
3.SuperDiamond
4.Difference of 2 Squares
*make sure to separate work to make your work
Back to
Problem
Solution
WORK:
_x2-1_
x2+x-2
(x-1)(x+1)
(x-1)(x+2)
Step 1) difference of 2
squares/diamond
Difference of 2 squares:
x2- 1
x x 1 1 (x-1)(x+1)
Diamond:
(x-1)(x+2)
-2
x
x
-1
+2
+1
Back to
Problem
Standard 12
Standard 13
Students add, subtract, multiply, and divide rational expressions
and functions. Students solve both computationally and
conceptually challenging problems by using these techniques.
Problem 82
Problem 83
Problem 85
Problem 84
Vocabulary
Rules &amp;
Strategies
82) 7z&sup2;+7z • z&sup2;-4
4z+8 z&sup3;+2z&sup2;+z
a. 7(z-2)
4(z+1)
c. 7z(z+1)
4(z+2)
Solution &amp;
b.7(z+2)
4(z-1)
d. 7z(z-1)
4(z+2)
Standard 13
Vocabulary
Factoring= The process of writing a number or
algebraic expression as a product.
GCF= For two or more numbers, the largest
whole number that divides evenly into each
number.
Back to
Problem
Rules &amp; Strategies
Check whether to use GCF, before choosing a
factoring method.
Divide out common binomials.
Back to
Problem
GCF:7z -&gt;
GCF: 4 -&gt;
7z&sup2;+7z • z&sup2;-4
&lt;- Difference of 2 Squares
4z+8
z&sup3;+2z&sup2;+z &lt;- GCF: z; Diamond
7z(z+1) • (z+2)(z-2)
4(z+2) z (z+1) (z+1)
7z(z+1) • (z-2)
4
z (z+1) (z+1)
Standard 13
= 7(z-2)
4(z+1)
Back to
Problem
Rules &amp;
Strategies
Vocabulary
2
84)
x+8x+16
x+3
2x+8
2
x-9
2
A 2(x+4)
2
(x-3)(x+3)
B 2(x+3)(x-3)
x+4
Solution &amp;
C
(x+4)(x-3)
2
2
D (x+4)(x-3)
2(x+3)
Standard 13
Vocabulary
Rational Expression: a quotient of polynomials
Back to
Problem
Rules &amp; Strategies
• Take the reciprocal of the fraction on the right
side of the sign and then multiply.
• Factor each term if needed.
• Divide out common factors, if needed.
Back to
Problem
Change to
multiplication and
take reciprocal of
second rational
expression
LOOK
BEFORE
YOU BEGIN!
2
x+8x+16
x+3
x+8x+16
W
+16
O
R
K
1x
+4
.
+
1x
+4
2
2
2x+8
x+8x+16
x-9 Difference of 2
2
x-9
x+3
2x+8 Greatest
Common
2
Factor: 2
2x+8
x-9
2x: 2 x
xx33
(x+4)(x+4) (x+3)(x-3)
8: 2 2 2 (x+3) (x-3) x+3
2(x+4)
2
GCF: 2
2x+8
2 2
+8
(x+4)
Solution
Diivide out!
(x+4) (x-3)
2
2(x+4)
Back to
Problem
Standard 13
Rules &amp;
Strategies
Vocabulary
85) Which fraction is equivalent to
3x
5
x
x
4+ 2
A.
Solution &amp;
X&sup2;
5
B. 9x&sup2;
20
c. 4
5
D. 9
5
Standard 13
Vocabulary
• Fraction: a ratio of two expressions or
numbers other than zero
• Equivalent: equal
Back to
Problem
Rules &amp; Strategies
•
•
•
•
•
Find the least common denominator (LCD)
Change division into multiplication
With the second fraction, take the reciprocal
Cross cancel
Multiply straight across
Back to
Problem
3x
5
x
x
4+ 2
x x
4=4
x 2x
+2=4
3x
4
3x
4
5
3x
find Common
denominator = 4
take reciprocal
4
5
Standard 13
Back to
Problem
Rules &amp;
Strategies
Vocabulary
83) Which fraction equals the product?
A 2x -3
3x+2
B 3x+2
4x-3
C x&sup2;-25
6x&sup2;-5x-6
D 2x&sup2;+7x-15
3X&sup2;-13X-10
Solution &amp;
Standard 13
Vocabulary
• Fraction = a ratio of algebraic quantities similarly
expressed
• Product = the answer of a multiplication problem
• Area Method = multiply outside add or subtract
inside terms (like terms)
• Numerator = the top numbers of a fraction
• Denominator = the bottom numbers of a
fraction
Binomials: two terms
Back to
Problem
Rules &amp; Strategies
•
•
•
Since there are two different fractions in parenthesis and both the numerators and
denominators are binomials you use “Area Method”
Area method is when you draw out a rectangle and divide it into 4 quadrants (4 squares)
Then, you put one binomial on the left side written out like this:
• Next, you put the other binomial adjacent
(next to) the fraction on top of the rectangle
• Then you multiply each term to each other and
Place each number in a square (label them 1, 2, 3, and 4)
• Write out numbers 1 and 4 into an expression as the first and
last terms : 2x&sup2; + 15
• Now, since numbers 2 and 3 are like terms combine them and put
them in between 2x&sup2; and +15 : 2x&sup2; +7x +15
• Repeat this process to the other binomial and place the expression
underneath 2x&sup2; +7x +15 .
Back to
Problem
Solution
+2x
2x&sup2;
10x
3x&sup2;
-15x
&sup2;
-3x
+2
+5
-5
-3
+3x
+x
+x
2x
-10
-15
3x&sup2;-13x-10
2x&sup2;+7x-15
Combine
2x&sup2;+7x-15 on the top and
3x&sup2;-13-10on the bottom
Back to
Problem
Standard 13
Standard 14
Students solve a quadratic equation by factoring or completing
the square.
Problem 57
Problem 60
Problem 62
Problem 58
Problem 61
Problem 59
Rules &amp;
Strategies
Vocabulary
59) What are the solutions for the
2
quadratic equation x + 6x = 16?
A -2,-8
B -2, 8
C 2,-8
D 2,8
Solution &amp;
Standard 14
Vocabulary
• Quadratic equation : an equation containing a
single variable of degree 2.
Back to
Problem
Rules &amp; Strategies
• the “diamond method”
• All the numbers have to be on one side of the
equation.
Back to
Problem
Solution
-16
x
+8
+6
ax 2+ bx + c = 0
x 2 +6x = 16
x
-16
-16
2
x2 + 6x – 16 =0
(x + 8) (x - 2) = 0 Solve each factor to = 0
x+8=0 x–2=0
- 8 - 8 +2 +2
x = -8
x=2
Back to
Problem
Standard 14
Rules &amp;
Strategies
Vocabulary
2
57) If x is added to x, the sum is 42.
Which of the following could be the
value of x?
A
B
C
D
Solution &amp;
-7
-6
14
42
Standard 14
Vocabulary
• Value: A number represented by a figure,
symbol; the value of x.
• Sum: a series of numbers to be added up
Back to
Problem
Rules &amp; Strategies
• Plug in the value of x
2
• Set up the expression (x) + x
Back to
Problem
Solution
Try
x=-7
2
(x) + x= 42
2
(-7) +(-7)= 42
49 + (-7)=42
42=42

Back to
Problem
Standard 14
Rules &amp;
Strategies
Vocabulary
62) What are the solutions for
2
A) 3
B) 3,-3
C) 1,-9
D) -1,9
Solution &amp;
Standard 14
Vocabulary
• Solutions - the answer itself
• Quadratic - involving the square and no higher
power of the unknown quantity; of the second
degree.
• Equation - an expression or a proposition,
often algebraic, asserting the equality of two
quantities.
Back to
Problem
Rules &amp; Strategies
•
•
•
•
First get into Quadratic Form ax2+bx+c=0.
Then use the diamond method
Finally separate the two factors to = 0.
Solve for “X”
Back to
Problem
Solution
ax2+bx+c=0
Step 1: x2-8x=9
-9 -9
Step 2: 1x2-8x-9=0
Diamond Method
-9
x
*
x
+1
-9
Step 3: (x-9) (x+1) = 0 Zero Product
X-9=0 x+1=0
Property
+9=9 -1=-1
x=9
x=-1
+
-8
Back to
Problem
Standard 14
Rules &amp;
Strategies
Vocabulary
58) What quantity should be added to
both sides of this equation to complete
the square?
x&sup2;-8x=5
A. 4
B. -4
C. 16
D. -16
Solution
Standard 14
Vocabulary
• Quantity: amount
Back to
Problem
Rules &amp; Strategies
• Identify “b”
• Plug in “b” to the equation: b &sup2;
2
• Solve to complete the square.
Back to
Problem
Solution
b=-8
-8 &sup2;
2
(-4)&sup2;
+16
Standard 14
Back to
Problem
Rules &amp;
Strategies
Vocabulary
60) Leanne correctly solved the equation x&sup2; + 4x = 6 by
completing the square. Which equation is part of her solution?
A ( x + 2 )&sup2; = 8
B ( x + 2)&sup2; = 10
C ( x + 4 )&sup2; = 10
D ( x + 4)&sup2; = 22
Solution &amp;
Standard 14
Vocabulary
• Completing the square : a method for solving
quadratic equations by using steps such as
1. Making sure you have the formula x&sup2; + bx + c
2. Find b &sup2;
2
3. Completing the square with the answer to
by adding it to both sides
4. Factor the left and simplify the right
5. Use Square root property then you’re
done
Back to
Problem
Rules &amp; Strategies
• Use the factoring method: completing the square
Back to
Problem
Solution
x&sup2; + 4x = 6
Completing the
Square
=
b = 4 = 2 &sup2; = 4 add to both sides of the equation
2
+ 4x + 4
=6+4
factor by diamond method
Perfect Square Trinomial :
X&sup2;
2
factor the right, simplify the left
Final Answer : ( x + 2 )&sup2; = 10 take square root of both sides
Back to
Problem
Standard 14
Vocabulary
Rules &amp;
Strategies
61) Carter is solving this equation by
factoring.
10x&sup2;-25x+15=0
Which expression could be one of his correct
factors?
a. x+3
b.x-3
c.2x+3
d. 2x-3
Solution &amp;
Standard 14
Vocabulary
Expression does not have an =
Factors can also be the binomials you get
when you diamond
Back to
Problem
Rules &amp; Strategies
• Ask yourself which factoring method do I use?
• GCF? Diamond? Super Diamond? Diff of 2
Squares
• Make sure to find two terms which can be
multiplied to the top, but added to the
bottom of the diamond.
• Keep asking yourself whether to factor more
or not.
Back to
Problem
GCF: 5
10x&sup2;-25x+15=0
5 5 5 5
+6
5 ( 2x&sup2; -5x +3) =0
5( x-1) (2x-3)=0
x = 2x
2x
-1
-2 -5 -3
These are all factors, but
one of them. Back to
Standard 14
Problem
Standard 15
Students apply algebraic techniques to solve rate problems, work
problems, and percent mixture problems.
Problem 85
Problem 86
Problem 88
Problem 89
Problem 90
Problem 91
Problem 87
Rules &amp;
Strategies
Vocabulary
87) Andy’s average driving speed for a 4-hour trip was 45
miles per hour. During the first 3 hours he drove 40 miles per
hour. What was his average speed for the last hour of his
trip?
A 50 miles per hour
B 60 miles per hour
C 65 miles per hour
D 70 miles per hour
Solution&amp;
Standard 15
Vocabulary
• Rate = A ratio that compares two quantities
measured in different units.
• Average = The sum of the values in a data set
divided by the number of data values. Also
called the mean.
Back to
Problem
Rules &amp; Strategies
.
• Set up formula distance = rate time
• Put the same units together
Back to
Problem
Solution
d = 45. 4
d = 180
180 = r . 3 isolate rate
3
3
60 = r
Back to
Problem
Standard 15
Rules &amp;
Strategies
Vocabulary
86) A pharmacist mixed some 10%-saline solution with
some 15%- saline solution to obtain 100mL of a
12%-saline solution. How much of the 10%-saline solution
did the pharmacist use in the mixture?
A)
B)
C)
D)
Solution &amp;
60mL
45mL
40mL
25mL
Standard
15
Vocabulary
Distributive Property: is when you
distribute what is outside of the
parentheses.
Isolate the variable: is to simplify the
equation using operations to get the
variable alone , on one side of the
equal sign.
Back to
Problem
Solution
x +y
12%
100
mL
10%
+
15%
=
12%
x
+
y
=
100mL
=
.12(100)=12
=100mL
.10x
+
x+y= 100 isolate y -&gt; y=(100-x)
0.10x+0.15y=12
0.10x+ 0.15(100-x)=12
0.10x+15-0.15x=12
-15
.15y
Work
-0.15x
+0.10x
-0.05x=-3
-0.05
-0.05x
-0.05
-15
0.10x-0.15x= -3
-0.05x=-3 isolate x
X=60mL
Standard 15
Back to
Problem
Rules &amp; Strategies
•
Make a drawing with the values
• Remember that you need to go two spaces to the left
when changing a value to a decimal
• Make a system of equation and use substitution
Back to
Problem
Rules and
Strategies
Vocabulary
88) One pipe can fill a tank in 20
minutes. While another pipe takes
30 minutes to fill the same tank.
How long would it take the two
pipes together to fill the tank ?
Solution and
a. 50min
b. 25min
c. 15min
d. 12min
Standard 15
Vocabulary
• None available
Back to
Problem
Rules &amp; Strategies
Find the number of minutes it takes to fill the
tank together.
There are 60 minutes in 1 hour
Put information into fraction (over 1) and find
the LCM.
Once LCM is found, distribute the LCM inside
whatever is in the parentheses .
Back to
Problem
will be found.
Solution
1
1
20m + 30m = 1
LCM:
LCM:
m=minutes
20: 22 5
Choose one with the greatest power
30: 2 5 3
22 . 5 . 3 = 60 Distribute 60 to original equation
60 1
1
20m + 30m = 1(60)
60
60
20m + 30m= 60 Reduce Fractions
3m + 2m= 60
5m= 60
5
5
m=12
Back to
Problem
Standard 15
Rules &amp;
Strategies
Vocabulary
89) Two airplanes left the same airport traveling in opposite
directions. If one airplane averages 400 miles per hour and
the other airplane averages 250 miles per hour, in how many
hours will the distance between the two planes be 1625
miles?
A 2.5
B 4
C 5
D 10.8
Solution &amp;
Standard 15
Vocabulary
• Distance= rate * time isolate time
• d= rt
r r
t= d
r
Back to
Problem
Rules &amp; Strategies
• Remember to add both planes’ averages
• Divide answer to total distance
Back to
Problem
Solution
1st plane
2nd plane
avg 400
avg 250
400 + 250 = 650 is the rate
time = 1625 (distance)
650(rate)
time= 2.5 hours
Back to
Problem
Standard 15
Rules &amp;
Strategies
Vocabulary
90) Lisa will make punch that is 25% fruit juice by adding
pure fruit juice to a 2-liter mixture that is 10% pure fruit
juice. How many liters of pure fruit juice does she need to
A.
B.
C.
D.
Solution
0.4 liter
0.5 liter
2 liters
8 liters
Standard 15
Vocabulary
• Pure=100% or .10
• Liters=amount
Back to
Problem
Rules &amp; Strategies
• Use substitution
• Goal: solve for y
• First, isolate “x” then plug it in the 2nd
equation
• NOTE: 2 liters of 10% is 2(.10)= 0.2
Back to
Problem
x= 25 % = .25 y= 100% because its it's PURE fruit juice = 1 Solve for y
Results in 2 liters of 10% = (.10) fruit juice
First, isolate x then plug it in.
x-y = 2
.25x - 1y = 2(.10)
x-y=2
(2.4) – y = 2
-2.4
-2.4
-y = -.4
-1 -1
-y=2-x
-1 -1 -1
y= -2+x
.25x-(-2+x)=.2
.25x+2-x=.2
-2 -2
.25x-x= -1.8
-.75x = -1.8
-.75= -.75
y= .4
x = 2.4
Standard 15
Back to
Problem
Rules &amp;
Strategies
Vocabulary
Miles
traveled
600
450
300
960
Gallons of
gasoline
20
15
10
x
91) Jenna's
car averaged 30 miles per gallon of gasoline on her trip.
What is the value of x in gallons of gasoline?
A
B
C
D
Solution &amp;
32
41
55
80
Standard 15
Vocabulary
• Average = The sum of all the values in a data set
divided by the number of data values. Also called
the mean.
• Value = The amount of ; or equivalency of .
• Rate = A ratio that compares two quantities
measured in different units.
• Ratio = A comparison of two numbers of
quantities .
• Equation = A mathematical statement that two
expressions equal.
Back to
Problem
Rules &amp; Strategies
• Rewrite the problem as an equation and solve
for ‘’X’’ ( 960 = 30x ) * The ration is 960 miles =
x number of gallons and you put the “30” in
because its miles per gallon.*
• When you have the equation “960 = 30x”
divide 30 on both sides to isolate “x”.
Back to
Problem
Solution
960 = 30x
30
30
x = 32
Back to
Problem
Standard 15
Standard 16
16. Students understand the concepts of a relation and a function, determine
whether a given relation defines a function, and give pertinent information about
given relations and functions.
Problem 92
Problem 93
Rules&amp;
Strategies
Vocabulary
92) Beth is two years older than Julio.
Gerald is twice as old as Beth.
Debra is twice as old as Gerald. The sum
of their ages is 38. How old is Beth?
A
B
C
D
Solution &amp;
3
5
6
8
Standard 16
Vocabulary
• Twice - Double
• Two years older - x +2
Back to
Problem
Rules &amp; Strategies
Given:
•
•
•
•
•
Find: Beth’s age
Beth is x
Julio is x-2
Gerald is 2x
Debra is 2(2x)
Total years of age is 38
them to
make the
expression
equivalent
to 38
Back to
Problem
Solution
x+(x-2) + 2x + 2(2x) = 38
x + x - 2 + 2x + 4x = 38
+2
+2
8x = 40
8x = 40
8
8
Beth is five years
x=5
Combine
like
terms
old.
Back to
Problem
Standard 16
Rules &amp;
Strategies
Vocabulary
93) Which relation is a function?
A.Input Output
B.Input Output
C.Input Output
D.Input Output
1
2
2
6
1
2
0
1
2
2
2
5
2
4
0
2
3
3
6
4
4
6
1
3
4
3
6
3
4
8
1
4
Solution &amp;
Standard 16
Vocabulary
• Relation = a property that associates two
quantities in a definite order, as equality or
inequality
• Function = set of ordered pairs in which none
of the first elements of the pairs appears twice
Back to
Problem
Rules &amp; Strategies
• Look @ input (x values) and the numbers in
the x value can not be repeated twice.
Back to
Problem
Solution
Input
Output
1
2
2
2
3
3
4
3
Back to
Problem
Standard 16
Standard 17
Students determine the domain of independent variables and the range of dependent variables
defined by a graph, a set of ordered pairs, or a symbolic expression.
Problem 94
Problem 95
Rules and
Strategies
Vocabulary
94) For which equation graphed below are all
the y-values negative?
a.
Solution and
b.
c.
d.
Standard 17
Vocabulary
• y-value ; (x, y)
Back to
Problem
Rules and Strategies
• Lines must cross at negative y-int. (Quadrant 2
&amp; 3) in order for all to be negative.
Back to
Problem
Lines cross
in the y-int
of -1.
4 1
3
2
Back to
Problem
Standard 17
Vocabulary
Rules &amp;
Strategies
95)What is the domain of the
function shown on the graph below?
A {-1,-2,-3,-4}
B {-1,-2,-4,-5}
C {1,2,3,4}
D {1,2,4,5}
Solution &amp;
Standard 17
Vocabulary
• Domain=the set of all first coordinates (xvalues) of a relation or function
• Function= A relation in which every domain
value is paired with exactly one range value
Back to
Problem
Rules &amp; Strategies
• Identify the ordered pairs (x,y)
• Put them in a t –chart
• Remember domain= x-values
Back to
Problem
*Only focus on
domain
Solution
X
y
1
-1
2
-2
4
-4
5
-5
Back to
Problem
Standard 17
Standard 18
18. Students determine whether a relation defined by a graph, a set of
ordered pairs, or symbolic expression is a function and justify the
conclusion.
Problem 96
96) Which of the following graphs represents a
relation that is not a function of x?
Vocabulary
Solution and
Rules and
Srategies
Standard 18
A
Back to
problem
B
C
D
vocabulary
1. Function_ : A relation ship or expression
involving one or more variables. (y=mx+b)
2. Vertical line test: putting strait lines going
through X Axis.
It tests to see
Whether it is a
function or
not.
Back to
problem
Solution
and
Remember
To
Give each
a vertical line test
D, isn`t a function
Because it is being
hit more than once.
Back to
problem
Standard
18
Rules and Strategies
You have to give it the vertical line
test in order to see if it is a function.
Rule: it is only a function if it only
hits the line once.
Back to
problem
Standard 19
Students know the quadratic formula and are familiar with its
proof by completing the square.
Problem 63
Problem 64
Rules &amp;
Strategies
Vocabulary
64) Four steps to derive the quadratic formula are
shown below. What is the correct order for these
steps?
Solution &amp;
A)
B)
C)
D)
I, IV, II, III
I, III, IV, II
II, IV, I, III
II, III, I, IV
Standard 19
Vocabulary
Derive= to reach or obtain by reasoning; deduce; infer.
Quadratic Formula= the formula for determining the
roots of a quadratic equation from its coefficients.
Quadratic= involving the square and no higher power
of the unknown quantity; of the second degree.
Formula= a general relationship, principle, or rule
stated, often as an equation, in the form of symbols.
Back to
Problem
Rules &amp; Strategies
This is the
order a
formula proof
should look
like.
Back to
Problem
Solution
Solution: A
Back to
Problem
Standard 19
Vocabulary
#63
Rules and
Srategies
Toni is solving this equation by completing the square.
Step1: ax2 + bx =-c
Step 2: x2 + ba X= - ca
Step 3:?
Which should be step 3 in the solution?
A)X2 =-cb - baX
B)X +ba = - cax
C)X 2 + baX+b2a = -ca + b2a D)X 2 +baX+ b2a 2 = - ca + b2a 2
Solution and
Standard 19
Vocabulary
1. Completing the square: An expression in the
form of x 2 + bx is not a perfect square. However,
you can use the relationship above to add a
term to x 2 + bx to form a trinomial that can be a
perfect square.
Back to problem
Step1: ax2 + bx =-c
b/a 2
Step 2: x2 + ba X= - ca 2
b x 1 2= b
Step 3:?
a
2
2a 2
x2 + ba X+ b2a 2= - ca +now add to both
b
sides.
2a 2
1) Do
completi
ng the
square
to both
sides
2) Now you
have your
Standard 19
Back to problem
Rules and Strategies
•First of you will need to divide b/a by 2 then you
have to find the square root. Then add or subtract
that number to both sides.
• You’re completing the square.
Back to problem
Standard 20
Students use the quadratic formula to find the roots of a seconddegree polynomial and to solve quadratic equations.
Problem 65
Problem 66
Problem 67
Problem 68
Rules &amp;
Strategies
Vocabulary
65) which is one of the solutions to the
equation 2x2-x-4=0 ?
A.
B.
C.
Solution &amp;
D.
1-√33
4
-1 +√33
4
1+√33
4
-1-√33
4
Standard 20
Vocabulary
• Equation: A mathematical statement that two
expressions are equal
Back to
Problem
Rules &amp; Strategies
• Identify a, b, and c
• Use the quadratic formula -b+ √b2-4ac to solve
this equation
2(a)
Back to
Problem
Solution
2x2-x-4
A=2
B=-1
C=-4
-b + √b2-4ac
1-4(-8)
2(a)
-b+ √(-1)-4(20(4)
1+32
2(2)
1+√33
33
4
1+√33
1-√33
4
4
Back to
Problem
Standard 20
Vocabulary
Rules and
Strategies
66.)Which statements best explains why
there is no real solution to the quadratic
equation 2x2 + x + 7 = 0?
A)The value of 12 – 4 ∙ 2 ∙ 7 is positive.
B)The value of 12 – 4 ∙ 2 ∙ 7 is equal to 0.
C)The value of 12 – 4 ∙ 2 ∙ 7 is negative.
D)The value of 12 – 4 ∙ 2 ∙ 7 is not a perfect square.
Solution and
Standard 20
Vocabulary
• Statement ; declaration of speech setting forth facts,
particulars, etc.
• Solution ; an explanation or answer
• Quadratic Equation ; ax 2 + bx + c = 0
Back to
Problem
Rules and Strategies
•
•
•
•
•
Identify a, b , and c in ax 2 + bx + c
Find the discriminant : b 2 – 4ac
X&gt;0 ; 2 solutions
X=0 ; 1 solution
X&lt;0 ; no solution
Back to
Problem
2x 2 + x + 7 = 0
Identify:
a= 2
b= 1
c= 7
Discriminant:
b 2 – 4ac
(1)2 – 4 ∙ 2 ∙ 7
=1 – 56
= - 55
Negative!
-55 &lt; 0 ;
No Solution
Back to
Problem
Standard 20
Rules &amp;
Strategies
Vocabulary
67) What is the solution set of the
A) -1
2,
Solution &amp;
1
4
B){-1+√2,-1 √2 }
C) -1+√7, -1-√7
8
8
D)No real solution
Standard 20
Vocabulary
• Solution: the process of determining
• Quadratic Equation: an equation containing a
single variable of degree 2. Its general form is
2
ax + bx + c = 0, where x is the variable and
a, b, and c are constants ( a ≠ 0).
Back to
Problem
Rules &amp; Strategies
•
•
•
•
Plug in the values into the Quadratic equation
Simplify if possible
Back to
Problem
Solution
X=-b&plusmn;√b2-4ac
2(a)
x=-2&plusmn;√4-32
16
x=-2&plusmn;√-28 Negative
16
=Φ
Back to
Problem
Standard 20
Rules &amp;
Strategies
Vocabulary
68). What are the solutions to the
equation
2
3x + 3 = 7x
A 7+ 85 or 7 - 85
6
6
B -7+ 85 or -7 - 85
6
6
C 7 + 13 or 7 - 13
6
6
D -7 + 13 or -7 - 13
6
Solution &amp;
6
Standard 20
Vocabulary
• Inequality = a mathematical statement that
two expressions are equal.
Back to
Problem
Rules &amp; Strategies
•
•
•
•
•
Plug in correctly
Take both sides of the square
Solve for “x”
2
+
Quadratic = -b b – 4ac
2a
Back to
Problem
Solution
3x2+ 3 = 7x Inverse operation to get in correct form
-7x
3x – 7x 2+ 3 = 0 a=3, b=-7, c=3
+ (-7)2 – 4(3)(3)
7
x=
2(3)
7 + 13
6
Back to
Problem
Standard 20
Standard 21
Students graph quadratic functions and know that their roots are
the x-intercepts.
Problem 69
Problem 70
Problem 72
Problem 71
Rules &amp;
Strategies
Vocabulary
69) The graph of the equation y=x&sup2;-3x-4 is shown below:
For what values or value of x is y=0?
Solution &amp;
A
B
C
D
x=-1 only
x=-4 only
x=-1 and x=4
x=1 and x=-4
Standard 21
Vocabulary
• Equation= two equally balanced expressions
• Value= quantity
Back to
Problem
Rules &amp; Strategies
• Identify x-intercept(s), root(s), zero(s),
solution(s)
Back to
Problem
Solution
Solution:-1,4
Make sure to go
from left to right.
Back to
Problem
Standard 21
Rules and
Strategies
Vocabulary
70) Which best represents the
graph of y= -x&sup2;+3?
a.
c.
Solution
b.
d.
Standard
21
Vocabulary
• Parabola = graph dervived from a quadratic
function.
Back to
Problem
Rules and Strategies
• After each step, make sure you graph before
moving onto the next one.
1) Axis of Symmetry
2) Vertex
3) Y-Intercept
4) Two Other Points
Back to
Problem
Axis of Symmetry = 0
B=0
-(0)
A= -1 2(-1)
Vertex = (0,3)
y= -(0)&sup2;+3
y=3
Y intercept= y = -x&sup2;+3
y= 3
2 Other Points
x= -1
x=-2
Y= -(-1)&sup2;+3 y = -(-2)&sup2;+3
Y= -1+3
y=-4+3
y= 2
y= -1
Back to
Problem
Standard
21
Rules &amp;
Strategies
Vocabulary
graphed has x-intercepts of
4 and -3?
A y= (x-3)(x+4)
B y= (x+3) (2x-8)
C y= (3x-1) (3x+1)
D y= (3x+1) (8x-2)
Solution &amp;
Standard 21
Vocabulary
• X-intercepts: x-coordinates of the point where
a graph intersects the x-axis.
• Quadratic function: A function that can be
written in the form of f(x)= ax+bx=c where a,
b, and c are real numbers and a zero.
Back to
Problem
Rules &amp; Strategies
• Do zero product property.
Back to
Problem
Solution
y= (x-3)(x+4)
y= (x+3)(2x-8)
(x-3) (x+4)=0
(x+3)(2x-8)=0
x-3=0 x+4=0
x+3=0 2x-8=0
+3 +3 -4 -4
-3 -3 +8 +8
x=3
x=-4
x=-3 2x=8
3,-4 -3,4
2 2
x=-3 x=4
-3,4 = -3,4
Back to
Problem
Standard 21
Rules &amp;
Strategies
Vocabulary
72) What are the real roots of the function in the graph?:
Solution &amp;
A
B
C
D
3
-6
-1 and 3
-6, -1, and 3
Standard 21
Vocabulary
• Roots: solution to function
Back to
Problem
Rules &amp; Strategies
• Look at the x-axis!
• Identify x-intercept(s), root(s), zero(s),
solution(s)
Back to
Problem
Solution
Solution:-1,3
Back to
Problem
Standard 21
Standard 22
Students use the quadratic formula or factoring techniques or both to
determine whether the graph of a quadratic function will intersect the xaxis in zero, one, or two points.
Problem 73
Rules and
Strategies
Vocabulary
73) How many times does the graph of
y = 2x2 – 2x + 3 intersect the x-axis?
a.
b.
c.
d.
Solution and
None
One
Two
Three
Standard 22
Vocabulary
• x-axis - the horizontal axis in a twodimensional coordinate system
• Discriminant- the name given to the
expression that appears under the square root
Back to
Problem
Rules and Strategies
• Try Super Diamond
• Try discriminant
Back to
Problem
Solution
y = 2x2 – 2x + 3
Try
:
2x2 – 2x + 3 = 0
TRY DISCRIMINANT doesn’t work
a=2
b2-4ac
b=-2
(-2)2-4(2)(3)
c=3
2x
+6
2x
-2
4-24
-20
No solution
Back to
Problem
Standard 22
Standard 23
physical problems, such as the motion of
an object under the force of gravity.
Problem 74
Problem 75
Problem 76
Rules &amp;
Strategies
Vocabulary
74) An object that is projected straight
downward with initial velocity v feet per second
travels a distance s=vt+16t2 , where t= time in
seconds. If Ramon is standing on a balcony 84 feet
above the ground and throws a penny straight
down with initial velocity of 10 feet per second, in
how many seconds will it reach the ground.
Solution &amp;
A) 2 seconds
B) 3 seconds
C) 6 seconds
D) 8 seconds
Standard 23
Rules &amp; Strategies
• Plug in 10 feet per second under v.
• Plug in 84 as s.
Back to
Problem
Vocabulary
• Velocity=the rate of speed with which
something happens; rapidity of action or
reaction.
Back to
Problem
Solution
S=vt+16t 2
84=10t+16t 2
-84
0=
-84
16t 2+10t-84
2
2
2
Gcf:2
2x2x2x2x3x7
8(-42)
0=2(8t 2 +5t-42)
8x
0=2(t-2)(8t+21) -16
8x
21
t-2=0 or 8t+21=0 5
t=2 or t= -218
You have to pick the one that’s most
1)Substitute known values
into equation.
2)Put it in standard form.
3)Solve by doing gcf and
then super diamond.
4)Now do zero property.
5) Find out which one makes
more sense because time
can`t be expressed as a
negative it has to be 2
seconds.
Back to
Problem
Standard 23
Rules &amp;
Strategies
Vocabulary
75)The height of a triangle is 4 inches greater
than twice its base. The area of the triangle is
168 square inches. What is the base of the
triangle?
A) 7 in.
B) 8 in.
C) 12 in.
D) 14 in.
Solution &amp;
Standard 23
Vocabulary
•
•
•
•
•
•
Height-The perpendicular distance from any vertex of a triangle to the side
opposite that vertex. Also called altitude. Sometimes the height is determined
OUTSIDE of the triangle.
Base:The side of a triangle to which an altitude is drawn. the base and the altitude
will be used to find the area
area-A = 1/2(bh), where b is the length of the base, and h is the length of the
altitude. A = Square root [s(sa)(sb)(sc)], where s is the semiperimeter and a, b, and
c are the lengths of the sides of the triangle.
triangle-a three-sided polygon
Altitude-segment from the vertex of a triangle perpendicular to the line containing
the opposite side.
Vertex-the point of intersection of lines or the point opposite the base of a figure
Back to
Problem
Rules &amp; Strategies
•
•
•
•
•
Find the height of the triangle
Plug in area for area formula
solve the formula
Use zero product property
FACTOR by GCF then Diamond !
Back to
Problem
-168
b
-12
Step: 6
x
b
+
+14
Solution
14
X+2
2
A=
168
12 168
Step: 2
168=1\2b(2b+4)
Step: 3
0=2(b-12)(b+14)
B-12=0
B=+12
B+14=0
B=-14
b
2(168)=2b2+4b
Step: 4
0=2b2+4b-336
Step: 5
Back to
Problem
Standard 23
Rules and
Strategies
Vocabulary
76) A rectangle has a diagonal that measures
10 centimeters and a length that is 2
centimeters longer than the width. What is
the width of the rectangle in centimeters?
a.
b.
c.
d.
Solution and
5
6
8
12
Standard 23
Vocabulary
• Diagonal - a line joining two nonconsecutive
vertices of a polygon or polyhedron
• Length – The measurement of the extent of
something from the vertical side
• Width - The measurement of the extent of
something from side to side
Back to
Problem
Rules and Strategies
1. Draw a diagonal in the rectangle
2. Use Pythagorean's theory to solve for the
Visual Picture
width.
A. a=2+w, w=width
B. b= w
C. c=diagonal, 10cm
10
2+w
3. (2+w)2+w2= 102
w
Back to
Problem
Solution
(2+w)2+w2 = 102
(2+w)(2+w)+w2=100
4+4w+w2+w2=100 Combine like terms and put in descending
-100
-100 order.
2w2+ 4w + -96 = Use super diamond method
(w+8)(w-6)= 0
-192
w= 2 w
2 w =w
8 16
-12 -6
Width = -8 or 6 width = 6
4
Back to
Problem
Standard 23
Standard 25.1
Students use properties of numbers to construct simple, valid
arguments (direct and indirect) for, or formulate counterexamples to,
claimed assertions.
Problem 23
Rules &amp;
Strategies
Vocabulary
23) John’s solution to an equation is shown below.
Given: x+5x+6=0
Step 1:
(x+2)(x+3)=0
Step 2: x+2=0 or x+3=0
Step 3:
x= -2 or x= -3
Which property of real numbers did john use for Step 2?
A
B
C
D
Solution &amp;
multiplication property of equality.
zero product property of multiplication.
commutative property of multiplication.
distributive property of multiplication over
Standard 25.1
Vocabulary
• Solution: the process of determining the
• Equation: A mathematical statement that two
expressions are equal.
• Zero Product Property: For real numbers p and
q, pq = 0 , then p= 0 or q =0 .
Back to
Problem
Rules &amp; Strategies
• Look at Step 2 and see which property it is.
• Remember which property is which and don’t
mix them up.
Back to
Problem
Solution
Step 2: x+2=0 or x+3=0
Zero product property of
multiplication
Back to
Problem
Standard 25.1
Rules &amp;
Strategies
Vocabulary
52) What is the perimeter of the figure shown
below, which is not drawn to scale?
X+13
3x
3x+2
8
2
X+5
Solution &amp;
Standard 10
Vocabulary
• Perimeter: sum of all sides
• Scale: size of the shape
Back to
Problem
Rules and Strategies
• Add ALL of the sides
• Combine like terms
Back to
Problem
• 3x+2 + x+ 13+3x+8+2+x+5
• 8x + 30
Back to
Problem
Standard 10
Rules &amp;
Strategies
Vocabulary
77) What is x2 – 4xy+ 4 y2 reduced to lowest terms?
3xy-6y2
A) x-2y
3
B) x-2y
3y
Solution &amp;
C) x+2y
3
D) x+2y
3y
Standard 12
Vocabulary
• Reduced: in simplest form
Back to
Problem
Rules and Strategies
• Look at numerator and decide which factoring
method is needed.
• Look at denominator and decide which
factoring method is needed.
• Divide out common factors.
Back to
Problem
x2 – 4xy+ 4 y2 (diamond)
3xy-6y2
( GCF)
(x-2y)(x-2y) divide out!!
3y (x-2y)
3y
+4
x
-2 y
-4
x
-2y
3xy -6y2
3y 3y
3y (x-2y)
Back to
Problem
Standard 12
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