Tutorial for the Algebra 2A/B Summer Assignment
Ex. 1 Solve  6  3x  9  27
+9
+9 +9
Add 9 to each section.
3 3x 36


3 3
3
Divide by 3.
1  x  12
Graph
< or > use an open circle when you graph
 or  use a closed circle
3
8x  12  3x  53x  6  2 x  7 Distribute 3 to (8 x  12) and
4
4
 5to(3x  6)
The result is 6 x  9  3x  15x  30  2 x  7
Combine Like Terms
Ex. 2 Solve
9 x  9  13x  37
 13x
 13x
22 x  9  37
+9
+9
22 x  28
Solve for x.
Divide by 22 and reduce if possible.
22 x  28

22
22
x
 14
11
Graph
Ex.3 Solve and Graph
4x  7  9
For greater than absolute value inequalities you need to create two inequalities. First copy the
original problem without the absolute value symbol. The second inequality is created by
reversing the inequality and taking the opposite sign of the # on the right. Your inequalities will
look as follows:
or
Solve the inequalities.
4x  7  9
4 x  7  9
4x  7  9
+7 +7
or
4 x  7  9
+7 +7
4 x  16
4 x  2
4 x 16

4
4
4x  2

4
4
x
x4
1
2
Graph
This is an or problem because the inequality is either > or  , which means when we graph the
solution we take the union of the two graphs.
Ex.4 Solve and graph 5x  6  9
+6 +6
or
3x  5  38
-5 -5
5 x 15

5
5
or
3x 33

3
3
x3
or
x  11
Graph
3x  2  8
Ex. 5 Solve and graph
You also need to create two inequalities for a less than absolute value inequality, but you will
write it as one single expression. To do this you must begin by writing the original inequality
without the absolute value symbol, than place the opposite sign of the # to the right in front of
the original inequality as shown below.
 8  3x  2  8
-2
-2 -2
Follow Ex.1 to help you solve the inequality.
 10 3x 6


3
3 3
Ex.6 Solve and graph
Graph
4 x  5  13
4 x  5  13
-5 -5
or
4x  8
x2
Follow the steps from Ex. 3.
4 x  5  13
-5
-5
4 x  18
or
x
 18
reduce
4
x
9
2
Graph
Ex.7 Solve 3x  2  17
This is an absolute value equation. You create two equations by writing the original equation
without the absolute value symbol, and the second equation is the same except you take the
opposite sign of the number on the right. See below.
3x  2  17
-2
-2
or
3x  2  17
-2
-2
3x  15
3x  19
3x 15

3
3
3x  19

3
3
 19
3
Follow the steps from example 7.
x
x5
Ex. 8 7  3x  28
7  3x  28
-7
-7
or
7  3x  28
-7
-7
3x  21
3x  35
3x 21

3
3
3x  35

3
3
x7
x
 35
3
For examples 9 – 11 write the equation of the line.
Ex.9 Given two points (5,3) and (-2,10), find the equation of a line.
y  y1
10  3
7
First find the slope given by the formula m  2
. m
=
 1
25 7
x 2  x1
Next use the slope-intercept form of an equation which is y  mx  b where m = the slope and b
= the y-intercept. Substitute in for x and y using any given point and substitute in for m from
step one to find b.
y  mx  b
3  1(5)  b
3  5  b
8b
Now substitute in for m and b into the equation y  mx  b. Answer: y   x  8
Ex.10 x-intercept is -4
y-intercept is 7
This is the point (-4,0).
This is the point (0,7).
Now find the slope and follow the steps in example 9.
m
70
7

0  (4) 4
y  mx  b
I chose to use point (0,7)
7
(0)  b
4
b7
7
Answer:
y
7
x7
4
Ex. 11 f (4)  3,
x
y
Point 1 (4,-3)
f (2)  5
x y
Point 2 (2,5)
Now find the slope and follow the steps in example 9. (Continued on the next page).
Now find the slope and follow the steps in example 9.
5  (3)
8
m

 4
24
2
y  mx  b
5  4 ( 2 )  b
5  8  b
13  b
Answer:
y  4 x  13
Ex.12 You want to fence in your yard. The area that you want to enclose is 800 square feet.
The length of the property is 40 feet. How much will it cost you to put a fence around your
property if fencing costs $35 a foot?
Area = length  width
A  lw
800  40 w
800 40 w

40
40
20  w
40
20
20
40
Now find the perimeter:
P  2l  2w
P  2(40)  2(20)
P  80  40  120 feet
The cost for fencing will be 120 feet  $35 = $4,200.
Ex.13 A plumber charges $65 to enter your home and $45 per hour of work.
a) Write an inequality that represents the possible number of hours the plumber could work for
$335.
$65 represents your fixed cost, $45x represents the number of hours worked times $45, and $335
represents the maximum cost, thus the following equation:
$65  45x  $335
b) Solve the inequality: 65  45x  335
-65
-65
45 x  270
45 x 270

45
45
Answer: x  6
Ex.14 The length of a soccer field can vary from 80 feet to 120 feet inclusive. Write an
inequality that describes the possible lengths of a soccer field.
Answer: 80  x  120
Ex.15 On your first five tests of the marking period, you earn grades of 95, 90, 80, 88, and 85.
What grade would you need to earn on your last test so that you have an average of 88.
Since you will have taken six tests and the average of the six tests is an 88, multiply 6  88.
This will give you the total points needed for an 88 average after 6 tests. Total points will equal
528 (6  88=528).
Now take the sum of the five tests and subtract that total from 528.
95+90+80+88+85=438
Total Points:
Minus the sum of the 5 scores:
528
-438
Score needed to average an 88:
90
Answer: You need a 90 on your last test to get an 88 average.
Ex. 16 In your own words, define what a function is. Then graph the function
f ( x)  mx  b
m = the slope =
is the same as
rise
run
f ( x) 
2
x 5.
3
y  mx  b
b = the y-intercept, which is the 1st point you plot on the y-axis.
See the next page for steps.
Step 1: Plot -5 on the y-axis.
Step 2: From -5, the slope tells you to go up two and right 3 where you will plot your 2nd point.
2
Connect the two points and you now have the graph of f ( x)  x  5
3
Ex.17 Graph the following inequality: y  5x  3
m  the slope =
rise  5

run
1
b  y  int ercept  3
Follow Ex. 16 for directions on how to graph y  5x  3.
Shade below the line because it is less than or equal to.  or  is a solid line because the points
on the line are included. < or > is a dashed line because the points on the line are not included.
Ex. 18 Write the equation of the line in slope-intercept form that passes through the points (7,-2)
and (3,10). Then graph the line.
Step 1: Find the slope. m 
y 2  y1 10  (2) 12

 3
x2  x1
37 4
Step 2: Now follow example 9 and use either point to find b.
y  mx  b
10  3(3)  b
10  9  b
b  19
Answer:  3x  19
Ex.19 Here is an example of a vertical line test.
a.
This is an example of a graph that passes
the vertical line test because when you
draw vertical lines they intersect only one
point.
b.
This graph fails the vertical line test
because when you draw vertical lines they
intersect in more than one point.
Ex.20 Graph the following absolute value function: y  2 x  2  3 . This is in the form of
rise
. To graph the function, start at point (0,0). H moves the
run
point (0,0) left or right. For ex., x  2 moves the graph 2 units to the right.
y  a x  h  k . a = the slope =
x  2 moves the point 2 units to the left. Wherever h lands on the x-axis, move up or down
according to what k is. 2 moves the point up 2 units and -2 down 2 units
y  2 x  2  3. From (0,0) move left 2, then down 3 and plot your first point at (-2,-3).
rise
. From the point (-2,-3) the slope tells you to rise 2 then plot 2 additional
run
points left 1 and right 1 to form a v figure. See the graph on the following page.
a = the slope =
Ex. 21 Write the absolute value function that has the following properties:
a) Reflects over the x-axis: this means the graph opens downward.
b) Has a vertical stretch of 3: this means the slope equals a =
3
.
1
c) Has a horizontal shift of 4: this means 4 the left from (0,0) and k = 4
d) Has a vertical shift up 4: from -4 on the x-axis move up 4 then plot the point (-4, 4)
Now substitute in the equation y  a x  h  k therefore y  3 x  (4)  4 . Now simplify to
get y  3 x  4  4 .
Ex.22 Evaluate the following function for x = -2:
f ( x)  4 x 2  3x  12
f (2)  4(2) 2  3(2)  12
f (2)  4(4)  6  12
f (2)  16  6  12
f (2)  2
Evaluate is another word for substituting in for x.
Follow the order of operations (PEMDAS).
Ex.23 At 4AM, there was 3 inches of snow on the ground and at 2PM there were 23 inches of
snow on the ground.
a) Write a model for this situation.
First, find the slope using:
m = (the difference in snowfall)  (the difference in time).
m
23  3
20inches

 2inches / hour.
2 PM  4 AM 10hours
The linear model is y  2 x  3 ; 2x represents the snowfall/hour and 3 represents the original
amount of snow.
b) Based on your model, how much snow was there at 4PM?
y  2(12)  3
From 4AM to 4PM is 12 hours and x = hours, therefore: y  24  3
y  27
Ex.24 Levi is getting stronger by lifting weights. Below is a table which represents the amount
of weight that Levi could lift each week.
Week
Amount
of Wt.
1
70
2
80
3
90
4
100
5
115
6
120
7
140
a) Make a scatter plot.
b) Are there any outliers?
c) Is there a positive or negative correlation? Explain the relationship between the week and
amount of weight.
d) Write the line of best fit to predict how much weight Levi lifted on the tenth day. Do by
hand.
a) To make a scatter plot let x = week and y == Amt. of Wt. Then plot the points
(1,70), (2,80), ( 3, 95),…
b) Outliers are points that don’t stay in the general form of a line.
c) If the graph is rising from left to right it is positive correlation.
d) Approximate the line of best fit in the form of y  mx  b.
Ex.25 Graph the line 5x  3 y  15. Determine the x- and y- intercepts. Also, determine the
slope of the function.
To find the x-intercepts, set y = 0, then solve for x.
5 x  3(0)  15
5 x  15
x3
Plot 3 on the x-axis.
To find the y-intercept, set x = 0, then solve for y.
5(0)  3 y  15
 3 y  15
y  5
Plot -5 on the y-axis, then connect the points to get your graph of a line.
Now to find the slope of the equation put the equation into slope-intercept form ( y  mx  b ) by
solving for y.
5 x  3 y  15
 3 y  5 x  15
5
y  x5
3
The slope is m 
5
3
Ex.26 Solve the following absolute value equation: y  4 x  7  12 for y  16.
16  4 x  7  12
First substitute 16 in for y.
28  4 x  7
7  x7
Then add 12 to both sides.
Then divide by 4.
Now, follow the steps from example 7.
x7  7
or
x  7  7
x = 14
or
x=0