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Algebra 1
Second Semester Exam Review
Chapter 6—Solving and Graphing Linear Inequalities
6.1 Solving One-Step Linear Inequalities
Graphing Linear Inequalities (one variable): Ex. x > 2
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Solving One-Step Inequalities: use the following Properties of Inequality
Addition Property of Inequality—when you add the same no. to both sides of the
inequality, the inequality remains the same (use with subtraction problems)
Subtraction Property of Inequality—when you subtract the same no. from both sides of an
inequality, the inequality remains the same (use with addition problems)
Multiplication and Division Properties of Inequality—when you multiply or divide both
sides of an inequality by the same positive no., the inequality remains the same;
when you multiply or divide both sides of an inequality with a negative number,
switch the inequality symbol.
6.2 Solving Multi-step Linear Inequalities
Use the methods from Section 6.1 until the inequality solved
Ex. 4x – 3 < 13,
4x < 16,
x<4
6.3 Solving Compound Inequalities
Compound Inequalities—two inequalities connected by and or or
Compound Inequalities with and: solve simultaneously, the answer is both answers
Ex. -2 < 3x – 8 < 10;
6 < 3x < 18;
2<x<6
(x is less than 6 and
greater than 2)
Compound Inequalities with or: solve separately, the answer is one or the other of the answers
Ex. 3x + 1 < 4 or 2x – 5 > 7;
3x < 3 or 2x > 12;
x < 1 or x > 6
(x is less
than 1 or greater than 6)
6.4 Solving Absolute Value Equations and Inequalities
Every absolute value equation has two answers, 1). When inside | | is positive, 2). When inside | |
is negative
Ex. | x – 2| = 5
“positive answer:” x – 2 = 5,
x=7
“negative answer:” x – 2 = – 5,
x=–3
Every absolute value inequality has two answers, 1). When in inside | | is positive, 2). When inside
| | is negative
Ex. |2x + 1| - 3 > 6
“positive answer:” 2x + 1 > 9,
2x > 8,
x>4
Move the -3 first, then split:
“negative answer:” 2x + 1 > – 9, 2x + 1 > – 10,
x<–5
6.5 Graphing Linear Inequalities in Two Variables
When graphing linear inequalities with 2 variables (x and y), graph with shading in (x, y)
coordinate plane
Ex. y > 2x + 1
Step #1: graph the corresponding linear equation Ex. graph y = 2x + 1
Step #2: use solid line for > or <; dashed line for > or <
Step #3: shade below the line for < or <; shade above the line for > or >
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6.6 Stem and Leaf Plots and Mean, Median, and Mode
Not covered in Algebra I
6.7 Box and Whisker Plots
Not covered in Algebra I
Chapter 7—Systems of Linear Equations and Inequalities
7.1 Solving Linear Systems by Graphing
System of Linear Equations—two equations with two variables
Solution of a System of Linear Equations—an ordered pair (x, y) that satisfies both of the
equations of a System of Linear Equations
Methods to find the Solution of a System of Linear Equations: Graphing, Substitution,
Elimination
Graphing Method: Graph the 2 equations of the System of Linear Equations. The point
where the 2 graphs intersect is the Solution of the System of Linear Equations
7.2 Solving Linear Systems by Substitution
Substitution Method: substitute one equation “into” the other to solve for x and y (the solution)
y  x 1
Ex.
substitute top equation into bottom: 2x + (x + 1) = -2
2 x  y  2
Solve for x: 3x + 1 = -2;
3x = -3;
x = -1
Substitute x = -1 into either of the equations to find y: y = -1 + 1 = 0
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7.3 Solving Linear Systems by Linear Combinations (Elimination)
Elimination Method: add the 2 linear equations to “eliminate” one variable, then solve for other
4 x  3 y  16
Ex.
Add both equations: 6x + 0y = 24; 6x = 24;
x=4
2 x  3y  8
Substitute x = 4 into either of the equations to find y: 4(4) + 3y = 16
16 + 3y = 16;
3y = 0;
y=0
Using multiplication first: you may have to multiply one or both of the equations by an
integer to set up a system of equations where adding the equations will eliminate a
variable
3x  5 y  6
12 x  20 y  24
Ex.
multiply top by 4, bottom by 3:
4 x  2 y  5
12 x  6 y  15
Now add: 0x + 26y = 39;
y = 1.5
x = -0.5
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S
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R
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7.4 Applications of Linear Systems
Use any of the three methods from Sections 7.1 to 7.3 to solve a system of linear equations
Tips: Use Graphing—to provide a visual model and to get an estimate of the solution
Use Substitution—if one of the equations will substitute easily into the other
Use Elimination—if the coefficient of x or y are opposites of each other (i.e. 3x and -3x)
7.5 Special Types of Linear Systems
There are 3 possibilities for the graphs of a System of Linear Equations:
The lines intersect (2 separate equations)—1 solution
The lines are parallel (2 equations with the same slope)—0 solutions
The lines coincide (2 similar solutions)—infinite solutions
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7.6 Solving System of Linear Inequalities
Use the techniques from section 6.5 to graph both inequalities—the solution is the overlap of the
shaded regions
Step #1 graph the corresponding linear equations
Step #2: use solid line for > or <; dashed line for > or <
Step #3: shade below each line for < or <; shade above the line for > or >
Step #4: final answer is the shaded area in common to both inequalities
Chapter 8—Exponents and Exponential Functions
8.1 Multiplication Properties of Exponents
Exponents—the number of times a base is multiplied by itself. Ex. 32 = 3 times itself twice (3∙3)
Multiplication Properties of Exponents: use these to simplify equations with exponents
Product of Powers Property—to multiply powers having the same base, add the exponents
am∙an = am + n
Ex. 32 ∙ 37 = 32 + 7 = 39
Power of a Power Property—to find a power of a power, multiply the exponents
(am)n = am ∙ n
Ex. (52)4 = 52 ∙ 4 = 58
Power of a Product Property—to find the power of a product, find the power of each factor
(a ∙ b)m = am ∙ bm
Ex. (2 ∙ 3)6 = 26 ∙ 36
8.2 Zero and Negative Exponents
Zero Exponent—a non-zero number to the zero power is 1: a0 = 1, if a ≠ 0
Ex. 40 =1
1
1
Negative Exponent—a-n is the reciprocal of an: a-n = n , if a ≠ 0
Ex. 3-5 = 5
a
3
8.3 Division Properties of Exponents
Division Properties of Exponents: use these to simplify equations with exponents
Quotient of Powers Property—to divide powers that have the same base, subtract the
am
37
mn
exponents
,
given
a
≠
0
Ex.

a
 3 7 5  3 2
an
35
Power of a Quotient Property—to find the power of a quotient, find the power of the
numerator and the power of the denominator and divide
m
am
a
   m , given b ≠ 0
b
b
3
43
4
Ex.    3
5
5
8.4 Scientific Notation
Scientific Notation—a number written in the form of c ∙ 10n, where 1 < c < 10, and n is an integer
Ex. 34,690 = 3.469 x 104
(move decimal 4 places left)
-4
Ex. 0.000722 = 7.22 x 10
(move decimal 4 places right)
Ex. 1.23 x 10-6 = 0.00000123
(move decimal 6 places left)
5
Ex. 4.9 x 10 = 490,000
(move decimal 5 places right)
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Chapter 9—Quadratic Equations and Functions
9.1 Solving Quadratic Equations by Finding Square Roots
Square Root—If b2 = a then b is a square root of a. Ex. If 32 = 9, then 3 is a square root of 9
Positive Square Roots—the square root that is a positive number. Ex. 9  3 , 3 is a
positive square root of 9
Negative Square Root—the square root that is a negative number. Ex.  9  3 , -3 is a
negative square root of 9
Positive and Negative Square Roots—when the problem is asking for both the negative
and positive square roots.
Ex.  9  3
Solving a Quadratic Equation: several methods will be taught over the next 2 chapters
Quadratic Equation in Standard Form—an equation in the form of ax2 + bx + c = 0
Solving x2 = d by Finding the Square Root:
If d > 0, then x2 = d has 2 real solutions, x = ± d
If d = 0, then x2 = d has 1 real solution, x = 0
If d < 0, then x2 = d has no real solutions.
Ex. x2 = 16, take square root of both sides,
x 2  16 , x ± 4
9.2 Simplifying Radicals
Properties of Radicals: use these to simplify problems with radicals
Product Property—the square root of a product equals the product of the square roots of
factors.
Ex. 4  100  4  100
ab  a  b
Quotient Property—the square root of a quotient equals the quotient of the square roots of
a
a
9
9
the numerator and denominator
,
Ex.


b
25
b
25
Simplifying Radical Expressions: continue simplifying radical expressions until none of the
following conditions are present:
1. No perfect square factors are in the radicand:
Ex. 8 can be simplified to 4  2  2 2
5
can be simplified to
16
3. No radicals appear in the denominator of a fraction:
1
3
3
1
Ex.
can be simplified to


3
3
3
3
2. No fractions in the radicand: Ex.
5
5

4
16
9.3 Graphing Quadratic Functions
Parabola—the graph of a quadratic equation
Vertex—the highest point on a downward opening parabola, or the lowest point on a upward
opening parabola
Axis of Symmetry—the line that passes through the vertex and divides the parabola into 2
symmetric parts
Graphing a Quadratic Equation ax2 + bx + c:
If a is positive, the parabola opens upward, if a is negative, the parabola opens downward
The vertex has an x-coordinate of – (b/2a)
The y-coordinate can be found by taking the value of x and subbing it into the
original equation and solving for y
The axis of symmetry is the vertical line, x = - (b/2a)
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9.4 Solving Quadratic Equations by Graphing
Solutions to a Quadratic Equation—the value(s) of x when the equation (y) = 0
Several Methods will be taught to find the solutions of a quadratic equation in the next 2
chapters
Graphing a quadratic equation to find the solutions:
Write the equation in the form ax2 + bx + c = 0
Write the related function, y = ax2 + bx + c
Graph this equation, the solutions (or roots) of this equation are the x-intercepts (the points
on the graph where y = 0)
9.5 Solving Quadratic Equations by the Quadratic Formula
Several Methods will be taught to find the solutions of a quadratic equation in the next 2 chapters
Quadratic Formula is another method to find the value(s) of x (the solutions to the
equation)
Quadratic Formula—the solutions of the quadratic equation, ax2 + bx + c = 0 are:
x
 b  b 2  4ac
, given a ≠ 0, and b2 – 4ac > 0
2a
2
Ex. For x + 9x + 14 = 0,
 9  9 2  4(1)(14)
,
x
2(1)
x = -2 and -7
9.6 Applications of the Discriminant
Discriminant—from the Quadratic Formula, b2 – 4ac is the Discriminant
The Number of Solutions of a Quadratic Equation:
If b2 – 4ac is positive, then the equation has 2 solutions
If b2 – 4ac is zero, then the equation has one solution
If b2 – 4ac is negative, then the equation has no real solutions
Chapter 10—Polynomials and Factoring
10.1 Adding and Subtracting Polynomials
Monomial—an expression with a constant, a variable, or a product of a constant and variable
Ex. 5x, 5x2, 5
Binomial—two monomials.
Ex. 5x + 5
5x2 + 5x
Polynomial—two or more monomials
Degree—the exponent of a monomial.
Ex. 5x4 has a degree of 4
Degree of a Polynomial—the largest degree of a polynomial.
Ex. 4x3 + 2x2, the
degree of the polynomial is 3
Standard Form—a polynomial written with the terms in descending order, from largest
degree to smallest degree.
Ex. 4x3 + 2x2 + 2x + 2
Adding Polynomials: group like terms (monomials with the same variable) and add the
coefficients. Ex. (5x3 + 4x2) + (2x3 – x2) = 7x3 + 3x2
Subtracting Polynomials: group like terms and add the opposite
Ex. (5x3 + 4x2) - (2x3 – x2) = (5x3 + 4x2) + (-2x3 + x2) = 3x3 + 5x2
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10.2 Multiplying Polynomials
FOIL—a multiplication method for two binomials, multiply the F(irst), O(uter), I(nner), and
L(ast) terms.
Ex. (3x + 4)(x + 5) = F(3x∙x) + O(3x∙5) + I(4∙x) + L(4∙5) = 3x2 + 15x + 4x + 20 =
3x2 + 19x + 20
Multiplying Polynomials with the Distributive Property: multiply first polynomial by each term in
the second polynomial. Use Distributive property twice to solve
Ex. (x + 3)(4x2 + 7x – 3) = (x + 3)(4x2) + (x + 3)(7x) + (x + 3)(-3) –1st Distributive Prop.
4x3 + 12x2 + 7x2 + 21x -3x – 9 –2nd Distrib. Prop.
= 4x3 + 19x2 + 18x – 9
10.3 Special Products of Polynomials
Special Product Patterns: binomial products that match these patterns follow these rules when
applying FOIL to the binomial factors:
Sum and Difference Pattern—(a + b)(a – b) = a2 – b2
Ex. (3x + 4)(3x – 4) = 9x2 – 16
Square of a Binomial Pattern—(a + b)2 = a2 + 2ab + b2
Ex. (x + 4)2 = x2 + 8x + 16
2
2
2
(a – b) = a – 2ab + b
Ex. (2x – 6)2 = 4x2 – 24x + 36
10.4 Solving Polynomial Equations in Factored Form
Using the Zero-Product Property: Method to solve quadratic equations
Zero-Product Property—If ab = 0, then a = 0 or b = 0
Steps: Set equation = 0, factor equation, use Zero-Product Property to find x
Ex. x2 + x – 6 = 0; (x – 2)(x + 3) = 0; x – 2 = 0 or x + 3 = 0; x = 2 or x = -3
Solutions to a Quadratic Equation: solutions are the values of x when y = 0 (the x-intercepts on
the graph of the quadratic equation
10.5 Factoring x2 + bx + c
Steps to Factor x2 + bx + c: find the pair of factors whose product is c, and whose sum is b
Ex. x2 – 5x + 6
the factors of +6 are: 1, 6; 2, 3; -1, -6; -2, -3
the factors of -2 and -3 add up to -5, so
(x – 3)(x – 2) is the factored form of x2 – 5x + 6
Hint: if the second sign is +, then both signs are either + or –
if the first sign is +, then both are +
if the first sign is -, then both are if the second sign is -, then one sign is + and one sign is –
if the first sign is +, then the “larger” number is +, “smaller” number is –
if the first sign is -, then the “larger” number is -, “smaller” number is +
10.6 Factoring ax2 + bx + c
Steps to Factor ax2 + bx + c: use guess and check until you find the correct factorization
The following pattern will help: ax2 + bx + c = (mx + p)(nx + q)
where
c = pq
and
a = mn
Ex. 2x2 + 11x + 5
factors for a are: 1, 2
factors for c are: 1, 5
try combinations of a and c until: (1x + 5)(2x + 1)
10.7 Factoring Special Products
Recognizing patterns in trinomials lead to easier factoring (the reverse of section 10.3)
Difference of Two Squares Pattern: a2 – b2 = (a + b)(a – b) Ex. 9x2 – 16 = (3x + 4)(3x – 4)
Perfect Square Trinomial Pattern:
a2 + 2ab + b2 = (a + b)2
Ex. x2 + 8x + 16 = (x + 4)2
a2 – 2ab + b2 = (a – b)2
Ex. x2 – 12x + 36 = (x – 6)2
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10.8 Factoring Using the Distributive Property
Factoring and the Distributive Property: factor out the Greatest Common Factor
Ex. 9x2 – 15x = (3 ∙ 3 ∙ x ∙ x) – (3 ∙ 5 ∙ x)
GCF is 3x, factor out 3x: 3x(3x – 5)
Complete Factorization: a polynomial is completely factored when all remaining factors are prime
Factoring by Grouping: factor a polynomial by grouping pairs of terms, then using the
Distributive Property twice
Ex. x3 + 2x2 + 3x + 6 = (x3 + 2x2) + (3x + 6) =
x2(x + 2) + 3(x + 2) = (1st Distributive Property)
(x2 + 3)(x + 2)
(2nd Distributive Property)
Chapter 11—Rational Equations and Functions
11.1 Ratio and Proportion
Ratio—a number written with a numerator and a denominator
Proportion—two equal ratios
a c
2 4
 , then 2 ∙ 6 = 3 ∙ 4
Cross Product Property—if  , then ad = bc
Ex.
b d
3 6
x4 x
 , then (x + 4) ∙ 5 = x ∙ 3, then 5x + 20 = 3x, then 2x = -20, then x = -10
Ex.
3
5
11.2 Percents
Verbal Model: Number Being Compared to Base is (×) a Percent of (=) Base Number
Ex. 21 feet is 30% of what number:
21 ∙ .30 = x
x = 70
21 30
is
%


Quick Model:
Ex. 21 is 30% of what number?
, x = 70
x 100
of 100
(use Cross Product Property to solve proportion)
11.3 Direct and Inverse Variation
Not covered in Algebra I
11.4 Simplifying Rational Expressions
Rational Number—any number that can be written with a numerator and denominator (ratio)
Rational Expression—any expression where the numerator and denominator can be written as
x2  3
polynomials.
Ex.
4x  7
Simplifying Rational Expression:
Step #1: factor numerator and denominators completely
Step #2: cancel factors that are common to both the numerator and denominators
Step #3: continue cancelling factors until the rational expression cannot be simplified any
longer (prime)
2 x 2  6 x 2 x( x  3) x  3
1
x 5
x 5
Ex.
=
=
Ex. 2
=
=
2
3x
2 x(3 x)
6x
x  8 x  15 ( x  5)( x  3) x  3
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Recognizing when Rational Expressions are Undefined: a rational expression is undefined when
the denominator = 0
To determine when a rational expression is undefined, set the denominator = 0, then solve
x 5
Ex.
, set (x – 3)(x + 2) = 0, thus x – 3 = 0 or x + 2 = 0, x = 3 or x = -2
( x  3)( x  2)
Therefore, this expression is undefined when x = 3 or when x = -2
11.5 Multiplying and Dividing Rational Expressions
Multiplying Rational Expressions: multiply numerators and denominators.
Ex.
3x 3 8 x 2
24 x 5
2

=
, simplify =
4
5
4 x 15 x
5
60 x
Dividing Rational Expressions: multiply by the reciprocal of the divisor:
Ex.
a c a d ad
   
b d b c bc
( x  3)( x  3) x  3
x2 9
x2 9
1

(
x

3
)
=
=
=

2
4x
4 x 2 ( x  3)
4 x 2 ( x  3)
4x 2
11.6 Adding and Subtracting Rational Expressions
Addition of Rational Expression with Like Denominators: add the numerators
Ex.
a c ac
 
b d bd
5 x5 5 x5

=
=1
x
x
x
a b ab
 
c c
c
Subtraction of Rational Expressions with Like Denominators: subtract numerators
a b a b
 
c c
c
4x
x2
4x  x  2

=
=1
3x  2 3x  2
3x  2
Addition and Subtraction of Rational Expressions with Unlike Denominators: find the Least
Common Denominator, re-write expression with the common denominator, then add or
subtract the numerators
Least Common Denominator (LCD)—the like denominator that is the least common
multiple of the original denominators
7
5
7(4 x )  5(3) 28 x  15
 2 , the LCD is 24x2, thus =
Ex.
=
6 x 8x
24 x 2
24 x 2
11
2
11(13)  2(6) 143  12 131

Ex.
, the LCD is 78x, thus =
=
=
6 x 13 x
78 x
78 x
78 x
Ex.
11.7 Dividing Polynomials
Not covered in Algebra I
11.8 Rational Equations and Functions
Not covered in Algebra I
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Chapter 12—Radicals and Connections to Geometry
12.4 Completing the Square
Completing the Square—to complete the square of the expression x2 + bx, add the square of half
2
the coefficient of x:
Ex. x2 + 10x = 24,
2
b
b 
x  bx      x  
2
2 
2
2
x + 10x + 52 = 24 + 5 ,
(x + 5)2 = 49
2
Solve for x: ( x  5) 2  49 ,
x + 5 = ±7,
x = -5 ± 7,
x = 2 and -12
Completing the Square is used to find the solutions of x from a Quadratic Equation by creating a
perfect square trinomial