Honors Algebra 2
Summer Assignment
2015-2016 Academic Year
Name:
Assignment Due: Thursday, September 3, 2015
Assignment Test: Friday, September 4, 2015
1
On the following pages are explanations, examples, and problems for you to complete. All of the topics covered are a
review of topics from previous math courses, and nothing requires new learning. Understanding these topics will be a
requirement for success in Honors Algebra Two. As a reminder, a graphing calculator is also required for Honors Algebra
Two. You should read the course description book for an explanation of which types of graphing calculators are
appropriate for the course.
I. Simplifying Algebraic Expressions
Explanation
An expression is a math sentence without an equal sign. Expressions can only be simplified.
An expression can be simplified by identifying and combine like terms.
To identify like terms, find two or more terms that have the same variables to the same powers.
To combine like terms, add their coefficients and maintain the variable portion.
Example
Simplify the following: 2 x  5 x 2  7 y  4 x  y  x 2
6 x  5x2  7 y  y  x2
Combine the terms with only x as the variable.
6x  6x  7 y  y
Combine the terms with only x2as the variable.
6 x  6 x2  8 y
Combine the terms with only y as the variable.
2
Problems
Simplify the following:
1] 3  2 y 2  7  5 x  4 y 3  6 x


3] 4 3 x  2 x 3  5  6 x
2] x 2  x  x 2  x
4]  4a  3b    9  10b    6a  7 
2
II. Solving Equations
Explanation
An equation is a math sentence with an equal sign. Equations can be solved.
If an equation has only one variable, the equation can be solved for that variable.
Isolate the variable using inverse operations, and find the value the variable takes on to make the equation true.
Example
Solve the following equation for x:
7  x  5
3
3
7  x
 53
3
 7  x   15
Multiply by 3 on both sides.
7  x  7  15  7
x  8
x 8

1 1
x  8
Subtract 7 on both sides
Simplify.
Simplify.
Divide by -1 on both sides.
Simplify.
Problems
Solve the following equations for the given variable.
5] 4  x  2  32
7] 4 x  18 
x
2
6]
x2
4
16
8] (2 x  4)  (5 x  1)  17
3
Explanation
If an equation has multiple variables, the equation can be solved for any single variable. To solve for a variable within a
variable-only equation, you must isolate the identified variable.
 Example
Solve for b in a 2  b 2  c 2 .
a 2  b2  c2
b2  c2  a 2
b2  c 2  a 2
b  c2  a2
To solve for b, isolate the b2 term by moving the a2 term.
The b2 is almost isolated, square root to eliminate the exponents.
The order of operations dictates that you square and subtract c and a
before you take the square root. Because you don’t know the values of
c and a you must stop. You can only take the square root of b2.
The equation is solved for b because b is completely isolated.
Problems
Solve the following equations for the given variable.
9] Solve for x:
xa
a
b
11] Solve for z: z 2  y  x
10] Solve for c:
a c

b d
12] Solve for x: b  x  a   c  d
4
III. Solving Inequalities
Explanation
An inequality is a math sentence with one of the following inequality signs: , , , or  .
Because inequalities have a range of values that represent their solution, we use a graph to represent those values. Just
like equations, solving an inequality begins with isolating the variable using inverse operations. Once the variable is
isolated you will use the chart below to graph a circle and line.
 Inequality 
Circle
x  ...
x  ...
x  ...
x  ...
Line
empty
points left
empty
points right
filled
points left
filled
points right
Some unique properties about inequalities follow:
~ The inequality sign will change direction from < to > or from ≤ to ≥ if while you are solving the following occurs:
1] you multiply or divide…
2] on both sides of the inequality…
3] by a negative number.
~Depending on where the variable is, a single inequality sign can be read two different ways:
x  3 is read as, “x is less than three,” OR “three is greater than x.”
5  y is read as, “five is greater than or equal to y,” OR “y is less than or equal to five.”
 Examples
Solve and graph the following: 2x  4  10
2x  4  10
2x  10
x  5
Subtract 4 from both sides to begin isolating x.
Divide by -2 on both sides to finish isolating x.
If you DIVIDE on BOTH sides by a NEGATIVE number, the
inequality sign must change direction (see Explanation).
The variable is isolated and ready to be graphed.
The inequality is read; “x is greater than negative five.”
x  5
The circle will be on -5.
The circle will be empty because the sign is not “or equal to.”
The arrow will be to the right because the sign is “greater than.”
–9
–8
–7
–6
–5
–4
–3
–2
–1
0
1
2
3
4
5
6
7
8
9
5
Problems
Solve and graph the following inequalities. You must draw your own number line.
14] 6  4 
13] 3x  5  17
15]
x6
2
3
x
2
16] 3(2 x  4)  0
IV. Solving Absolute Value Equations and Inequalitities
Explanation
Absolute value is defined as a number’s distance from zero on a number line.
Distance is defined such that it is always positive: d 
 x1  x2    y1  y2 
2
2
.
Because distance is always positive, Absolute Value always PRODUCES positive numbers. However, its inputs can
be any number at all. Consider the following examples:
The 5 is equal to 5.
The 5 is also equal to 5.
Although the output is the same in both examples, the inputs are different. Because we are looking for inputs
when we solve an absolute value equation, there will inevitably be two solutions. So to solve an absolute value
equation, you must isolate the absolute value expression, and then split the equation into two possible solution
paths; one positive and one negative.
 Examples
Solve: 2 x  3  1  9
2 x  3 1  9
Add 1 to both sides to being isolating absolute value expression in the equation.
2 x  3  10
Divide both sides by 2 to fully isolate the absolute value expression.
x3  5
The absolute value expression is isolated, so split the equation into a positive
x 3  5
x2
x  3  5
x  8
path and a negative path. This will generate two solutions.
Solve both equations independently by subtracting three on both sides.
The two solutions that would make the equation true are 2 and -8.
6
Problems
Solve the following equations.
17] 3 x  5  21
19]
18] 19  3  4 x  1
2x  3
 11
5
20] 27  4
x
2
5 9
Explanation
An Absolute Value Inequality is a math sentence with an absolute value expression and an inequality sign. To solve an
absolute value inequality, you will still isolate the absolute value expression first. Once Isolated, you will create two
inequalities to solve, and then you will graph both solutions on one number line.
~If the absolute value expression is >  other side  , then setup up two inequalities like this:
 absolute value expression    other side 
 absolute value expression    other side 
~If the absolute value expression is <  other side  , then setup up two inequalities like this:
  other side    absolute value expression 
 absolute value expression    other side 
 Examples
Solve: 2 x  3  10
x3  5
5  x  3
x 3 5
8  x
x2
–9
–8
–7
–6
–5
–4
–3
–2
–1
0
Divide by 2 on both sides to isolate the absolute value expression.
Split the equation into two paths using the setup given above.
Graph both solutions on one number line.
1
2
3
4
5
6
7
8
9
7
Problems
Solve and graph the following inequalities.
21] 3 x  5  21
23] 1  5
x
6
2
22] x  1  4  14
24] 8  4 x  2
V. Properties of Exponents
Explanation
There are six properties of exponents you need to know before studying Algebra 2. More properties will be introduced
in Algebra 2, but they build on the basic properties below. Examine the properties and their examples.
 Examples
Product Property
Quotient Property
a x  a y  a x y
ax
 a x y
ay
Example: 53  59  512
Power to a Power Property
a 
x y
a
Example:
Product to a Power Property
 ab 
x y
  6
Example: 6
4 5
108
 104
1012
x
 a xb x

20
Example: 19xy
 19 x y
2 3
Zero Exponent Property
Negative Exponent Property
a 0  1 except a  0
an 
Example: 7 0  1
3
3
6
1
an
Example: 4 3 
1
5
also
 5x 2
3
2
4
x
8
Problems
Simplify the following expressions by combining all similar bases, and removing any negative or zero exponents. You
may leave constants to a power as they are.
25] 2 x 2 y 4  7 xy 3
 82 x 2 y 7 z 
27] 
5 9 
 12 y z 

2
29] 7x y

3 4
0
26]
4 x2 y3
2 xy8
9 x 2 y
28]
3 z 4
30]
3a 2b3  4ac 4
12a 2bc3
9
VI. Distributing
Explanation
Distributing is the act of taking one or more terms, and multiplying them by another grouping of one or more terms.
Some teachers call this FOIL, but that expression is very limited and does not always work. Keep in mind that distributing
is very simple already; it means multiply everything in one group, by everything in another group.
 Examples
Distribute and simplify the following
( x  5)
x  5
Here, a negative is outside the quantity. That is the same as -1.
Multiply the -1 by both terms inside and remove the parenthesis.
 x  3 x  2
Here there are two terms in each binomial.
 x  3 x  2
Take the first term in the first binomial and multiply it by both terms in the second.
x2  2 x
 x  3 x  2
The results are listed on this line.
Take the second term in the first binomial and multiply it by both terms in the second.
x 2  2 x  3x  6
x2  x  6
The results are listed on this line.
Combine any like terms and you are done.
Problems
Distribute the following expressions, and combine any like terms that result.
31] 5  x  3
33]
 x  1 x  6
35]  3x  2  y  9
32] 6 x   x  12 y 

2
34] 4 x  2

  x  1
2
36]  x  1 x  3x  2

10
VII. Linear Forms
Explanation
A linear equation is an equation with the x term to the power of one. There are several forms of linear equations, each
with its own advantages. Because each form has slightly different variables, it’s important to know the one common idea
behind each form: when fully simplified there should be a value (number) in for every variable except for x and y.
Slope-Intercept Form
Point-Slope Form
y  y1  m  x  x1 
y  mx  b
m  slope

b  y-intercept

m  slope


 x1 , y1   a point
Standard Form
Slope equation
m
Ax  By  C
 A, B, C 
y2  y1
x2  x1
m  slope

 x1 , y1   a point

 x2 , y2   another point
numbers
 Example
Write an equation of a line that goes through the points  3, 4 ,  5, 8 .
y  y1  m  x  x1 
We’ll use point-slope form since we already have a point.
m
 8     4 
 5    3 
We just need to find slope, so we use the slope equation.
m
4 1

8 2
Simplify to find the slope.
1
y   4      x   3 
2
y  4  12  x  3
Substitute a point and the slope into point-slope form.
Simplify and you are done.
Problems
37] Write an equation of a line that goes through the points  2,0  ,  3,1 .
38] Write an equation of a line that goes through the point  8, 6  and has slope of
3
4
in slope-intercept form.
39] Find the x and y intercepts of the equation 4 x  3 y  12
40] Rewrite y  9 
5
3
(3x  6) in standard form.
11
VIII. Graphing
Explanation
When graphing a line you need at least two points. You’ll notice from the previous section that you also need to points
to establish the slope of a line. Slope is a number that represents how steep a line is. The bigger the number is for slope,
the steeper the line is. Slope can be positive, negative, zero, or undefined. Examples of each are given below.
Parallel lines relate to slope as well. Parallel lines have the same slope, but different y-intercepts. Perpendicular lines
(lines that form right angles) have opposite and reciprocal slopes.
 Examples
Positive Slope (m = 4)
Negative Slope (m = -½ )
 y  2x 1
Parallel 
 y  2x  3
Undefined Slope (x = 3)
Zero Slope (y = 1)
 y  3x  3
Perpindicular 
1
y   3 x  2






12
Problems
41] Write an equation of a vertical line containing the point  4,5  .
42] Write an equation of a horizontal line containing the point  3, 1 .
43] Write an equation of a line parallel to the line y  14 x  9
.
44] Write an equation of a line perpendicular to the line y  5 x
IX. Systems of Linear Equations
Explanation
A system of equations is the combination of two or more equations into a single problem.
When you graph an equation, you graph its points. These points represent solutions. So a graph is all possible solutions
to an equation. When you solve a system of equations you are trying to find solutions to all the equations in the system,
so you want points that are shared by all the graphs. Below is an illustration of the types of solution you can get when
solving a system, and a general explanation for one method for solving systems of equations: Graphing.

 Examples
Zero Solutions
One Solution
Infinite Solutions
y  x

y  x  2
 y  x 1

 y  x  3
 y  x 1

3 y  3 x  3
There are zero solutions to this
system because the lines are
parallel, and parallel lines do
not share any points.
There is one solution to this
system because the lines are
intersect, and they share
exactly one point  2,1 .
There are infinitely many
solutions to this system
because there are actually two
lines on this graph. The lines
are exactly the same therefore
they share all their points!
13
Problems
Graph the following systems to find the number of solutions that exist.
 y  23 x  1
45] 
y 
6
9
x 1
 y  2x 1
 y  4 x  5
46] 
 y  42 x  3
3 y  6 x  9
47] 


Explanation
Another method for solving systems of equations is named Substitution. Here your steps to solve are:
1) Isolate a variable in one of the equations.
2) Substitute the expression the isolated variable is equal to, into the other equation.
3) Solve for the only variable left in the other equation.
4) Use the value you found in Step #3 to solve for the remaining variable.



 Example
4 x  3 y  4
2 x  y  7
Solve the following: 
2x  7  y
2x  7  y
Add y across in the second equation to begin isolating it
Subtract 7 across and y is isolated. Step one is complete.
4x  3 2x  7  4
Substitute the expression y is equal to into the other equation. Step two is done.
4x  6x  21  4
10x  25
x  2.5
Begin to solve for x.
Continue to solve for x.
Now x is isolated and solved for. Step three is complete.
2  2.5  7  y
Substitute the value of x into one of the original equations to solve for y.
2  y
Solve for y, your solution is  2.5, 2  , and the graphs cross at that point.
14
Problems
Solve the following systems using the substitution method.
 y  6 x  11
2 x  3 y  7
49] 
7 x  2 y  13
 x  2 y  11
51] 
48] 
50] 
4 x  3 y  18
 y  2
2 x  6 y  6
7 x  8 y  5
Explanation
The third and last method for solving systems is named Elimination. In this method we use the Addition Property of
Equality from Geometry to add one equation to the other, thus allowing a term to be eliminated.
1) Use multiplication to make one term in each equation become the exact opposite of the other.
2) Add respective sides of the equations to eliminate a variable.
3) Solve for the only variable left in the equation.
4) Use the value you found in Step #3 to solve for the remaining variable.

Example
4 x  3 y  4
2 x  y  7
Solve the following: 
2  2 x  y    7   2 Multiply both sides by -2 so that the x terms are exact opposites in the system.
4 x  2 y  14
 4x  2 y   4x  3 y  4   14
Simplify and step 1 is done.
Add the left sides and right sides of the equations respectively.
5 y  10
Simplify and the x terms are eliminated. Step 2 is done.
y  2
Solve for y. Step 3 is done.
2 x   2  7
Substitute the value of y into one of the original equations to solve for x.
x  2.5
Solve for x, your solution is  2.5, 2  , and the graphs cross at that point.
15
Problems
Solve the following systems using the elimination method.
4 x  2 y  12
4 x  8 y  24
53] 
 x  y  11
2 x  y  19
4 x  9 y  9
 x  3 y  6
55] 
52] 
7 x  y  19
2 x  3 y  19
54] 
X. Simplest Radical Form
Explanation
Radical
A radical is the square root symbol, and the radicand is the expression under the radical.
x
Radicand
Simplest Radical form is achieved when (1) all perfect squares have been removed from the radicand, (2) there are no
radicals in the denominator, and (3) there are no fractional radicands.
Examples
8
20
4 ⋅ 2
4 ⋅ 5
2 · 2
2 · 5
x3
48
16 ⋅
3
4 · 3
x2 ⋅ x
x ⋅
x
Problems
Write the following expressions in simplest radical form.
56]
32
57]
900
58]
a 2b 4
59]
8x 2 y 3 z 6
16
Example
1
2

1
2
2


2
2
2
Problems
Rationalize the denominators of the following radical expressions.
60]
16
7
 Examples
61]
140
6
3
5

3
5
15


5
5
5
5
8

5
8
40 2 10
10




8
8
4
8
8

Problems
Write the following expressions in simplest radical form.
62]
50
100
63]
20
60
XI. Functions
Explanation
A relation is a rule that assigns any input to any output. We list matched inputs and outputs as points. The
following is an example of a small relation: 1, 2  , 1,5 ,  4,7  ,  0,9  . In that relation 1 is matched with 2 and
with 5, 4 is matched with 7, and 0 is matched with 9. Relations have no restrictions; they match anything to anything.
A function is more specific than a relation. A function has one rule: each input can have only one output.
The last relation could not be a function because the input 1 has two outputs of 2 and 5. However, the following relation
is a function because it satisfies the rule:
1, 2 , 3,5 ,  4,7  , 0,9 . With a 3 in place of the second one, every single
input has only one output.
On a graph, all functions pass a visual test we call the Vertical Line Test. If you slide a vertical line from left to right along
the graph, it should only touch one part of the graph at a time. If it touches two or more parts simultaneously, the
relation is not a function.
17
Examples
 2,3 ,  4,6 ,  8,12 ,  10, 24 This relation is a function; each input has only one output.
This is a function because a
This is not a function because a
vertical line would touch two
points on the graph if placed
behind the left most point.

vertical line would only touch

one point on the graph no

matter where it is placed.


Problems
Identify if the following relations and graphs are functions.
64]  2,3 ,  4,8 ,  6,10  ,  2,14 

66]

65]
 1,5 , 0,10 , 1,15, 2, 20 
67]
18