2x+3y=6
-2 x
-2x
3y=6-2x
Chapter 3, Math 065
Disclaimer: This file is a compilation of problems created by Math 065 instructors at WWCC. These problems are meant only to give students extra practice in preparation for exams or for understanding a particular
topic. We do not promise that the problems below are problems that will appear on your exam in Math 065.
1.
3y
6-2 x
ÅÅÅÅ
ÅÅÅÅ = ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ
3
3
y = 2 - ÅÅÅÅ23 x
Plot the points AH4, 5L, BH0, -3L, CH-1, 4L, DH5, -2L, and EH-4, -2L. Be sure to label each point.
-2
m = ÅÅÅÅ
ÅÅÅÅ
3
Answer:
6x-4y=8
-6 x
-6x
-4 y = 8 - 6 x
A
C
4
-4 y
8-6 x
ÅÅÅÅÅÅÅÅ
ÅÅ = ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ
-4
-4
2
y = -2 + ÅÅÅÅ32 x
-4
-2
2
E
3
m = ÅÅÅÅ
2
4
D
-2
B
2.
Given the line, 4 x - 8 y = 24, state the y-intercept and the slope and then graph the line.
4 x - 8 y = 24
-4 x
-4x
-8 y = 24 - 4 x
-8 y
So the lines are perpendicular (their slopes are negative reciprocals).
4.
Write the equation of the line passing through H4, -5L and H3, 7L.
The
slope
of
the
line
is:
y2 -y1
7-H-5L
12
m = ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅ = ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅÅ = ÅÅÅÅ
ÅÅÅÅ = -12
x -x
3-4
-1
2
24-4 x
ÅÅÅÅÅÅÅÅ
ÅÅ = ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅ
-8
-8
y = -12 x + b
7 = -12 H3L + b
7 = -36 + b
+36 + 36
43 = b
y = -3 + ÅÅÅÅ12 x
y = ÅÅÅÅ12 x - 3
The y-intercept is: b = -3, the slope is m = ÅÅÅÅ12
2
The equation of the line is: y = -12 x + 43
1
-2
5.
2
4
6
8
10
Given the function f HxL = » 5 - 2 x2 » +3 x - 1, find:
HaL
f H0L
HbL
f H-1L
-1
-2
-3
-4
3.
1
Using y = mx + b:
Determine whether the pair of lines is parallel, perpendicular, or neither:
2x+3y= 6
6x-4y= 8
f H0L = » 5 - 2 H0L2 » +3 H0L - 1
= » 5 - 2 H0L » +0 - 1
= » 5 - 0 » -1
= » 5 » -1
= 5-1
=4
HcL
6.
f H-1L = » 5 - 2 H-1L2 » +3 H-1L - 1
= » 5 - 2 H1L » -3 - 1
= » 5 - 2 » -4
= » 3 » -4
= 3-4
= -1
f H2L
This is not linear
HbL
2 Hx - 7L + 1 = 8 y - 3
This is linear
8.
Find the x and y intercepts for the following graph.
5
4
f H2L = » 5 - 2 H2L2 » +3 H2L - 1
= » 5 - 2 H4L » +6 - 1
= » 5 - 8 » +5
= » -3 » +5
=3+5
=8
HdL
f HtL
HeL
f H3 - xL
3
2
1
-5
-4
-3
f HtL = » 5 - 2 HtL2 » +3 HtL - 1
f H3 - xL = » 5 - 2 H3 - xL2 » +3 H3 - xL - 1
0
2 H0L - 6 = -6
9.
4
5
Which of the following points is a solution to the linear equation 3 x - 4 y = 8?
HaL
HbL
2 H1L2 - 6 = -4
H0, 2L
(4,1)
2 H2L2 - 6 = 2
-1
Yes, H0, 2L is a solution to 3x-4y=8
1
2
-2
10.
-6
Determine whether the following are linear equations of two variables.
-4 x2 + 3 y = 9
3 H0L - 4 H2L = 8 ?
0 - 8 = 8?
-8 = 8 ?
3 H4L - 4 H1L = 8 ?
12 - 4 = 8 ?
8 = 8?
Determine if the following ordered pairs are solutions to the equation 2 x - 3 y = 7.
HaL
H5, 1L
-4
HaL
3
The x-intercepts are: H-2, 0Land H2, 0L. The y-intercept is: H0, -4L.
2
2
7.
2
-4
No, H0, 2L is not a solution to 3x-4y=8
-1 2 H-1L2 - 6 = -4
-2
1
-3
y
2 H-2L2 - 6 = 2
2
-1
-5
-2
1
-1
-2
Use a T-chart to graph y = 2 x2 - 6
x
-2
Yes, H5, 1L is a solution to 2 x - 3 y = 7
HbL
(2,-1)
2 H5L - 3 H1L = 7 ?
10 - 3 = 7 ?
7 = 7?
2 H2L - 3 H-1L = 7 ?
4 - H-3L = 7 ?
7 = 7?
Yes, H2, -1L is a solution to 2 x - 3 y = 7
HcL
-5 y
30
ÅÅÅÅÅÅÅÅ
ÅÅ = ÅÅÅÅ
ÅÅÅÅ
-5
-5
(-2,-4)
No, H-2, -4L is not a solution to 2 x - 3 y = 7
11.
3 H0L - 5 y = 30
0 - 5 y = 30
-5 y = 30
2 H-2L - 3 H-4L = 7?
-4 - H-12L = 7?
8 = 7?
y = -6
y - intercept = H0, -6L
13.
Sketch the graph of each linear equation given.
HaL
x = 17
This is a vertical line:
Sketch the graph of the equation -5 x - 2 y = 6.
-5 x - 2 y = 6
+5 x
+5x
-2 y = 6 + 5 x
-2 y
6+5 x
ÅÅÅÅÅÅÅÅ
ÅÅ = ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ
-2
-2
y = -3 So, m =
-5
ÅÅÅÅ
ÅÅÅÅ
2
17
ÅÅÅÅ52
x
and b = -3
HbL
4
2
-3
-2
-1
y = -6
This is a horizontal line:
1
2
-2
-4
-6
-8
12.
Find the x and y intercepts for the graph of the linear equation 3 x - 5 y = 30.
x-intercept (y = 0):
3 x - 5 H0L = 30
3 x - 0 = 30
3 x = 30
3x
30
ÅÅÅÅ
ÅÅÅÅ = ÅÅÅÅ
ÅÅ
3
3
x = 10
x - intercept = H10, 0L
y-intercept (x = 0):
-6
14.
For each set of graphs, determine which line has the greater slope (give reasons for your answers).
HaL
y = ÅÅÅÅ32 x + b
1 = ÅÅÅÅ32 H-1L + b
10
Line 1
7.5
5
2.5
-10
-5
5
-2.5
1 = - ÅÅÅÅ32 + b
Line 2
10
+ ÅÅÅÅ32 + ÅÅÅÅ32
ÅÅÅÅ52
-5
-7.5
The equation of the line is: y =
-10
Line 1 has a greater slope because it is steeper and increasing.
HbL
17.
3
ÅÅÅÅ
2
x+
=b
ÅÅÅÅ52
Use the slope and y-intercept to sketch a graph of the line 5 x - 2 y = 6.
We want to get the equation into the form y = mx + b
10
Line 2
5
Line 1
2.5
-10
-5
5x-2y= 6
-5 x
-5x
-2 y = 6 - 5 x
7.5
5
-2.5
-2 y
6-5 x
ÅÅÅÅÅÅÅÅ
ÅÅ = ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ
-2
-2
10
y = -3 + ÅÅÅÅ52 x
-5
-7.5
m = ÅÅÅÅ52 , b = -3
-10
10
Line 1 has a greater slope because Line 2 is decreasing so it has a negative slope and Line 1 is increasing so it has
a positive slope.
15.
5
For each set of points, find the slope of the line passing through the points, then find its equation.
HaL
H-4, -3L, H52, -3L
The
slope
-4
HbL
H3, 4L, H3, -8L
-15
18.
slope
is:
y2 -y1
4-H-8L
m = ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅ = ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅÅ
x -x
3-3
2
Determine whether the following lines are parallel, perpendicular, or neither.
HaL
Find the equation of the line passing through the points H-3, -2L and H-1, 1L.
2x-4y=5
-2 x
-2x
-4 y = 5 - 2 x
-4 y
2y
1+x
ÅÅÅÅ2ÅÅÅÅ = ÅÅÅÅ
ÅÅÅÅÅ
2
y=
y2 -y1
1-H-2L
3
m = ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅ = ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅÅÅÅÅ
x2 -x1
-1-H-3L = ÅÅÅÅ
2
Using y = mx + b :
The first line is -x + 2 y = 7 and the second line is 2 x - 4 y = 5.
-x + 2 y = 7
+x
+x
2y= 7+x
1
The slope is undefined so it is a vertical line. The equation of the line is: x = 3
16.
4
-10
1
The slope is m = 0 so it is a horizontal line. The equation of the line is: y = -3
The
2
-5
is:
y2 -y1
-3-H-3L
m = ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅ = ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅÅÅÅÅ = 0
x -x
52-H-4L
2
-2
ÅÅÅÅ12
+
5-2 x
ÅÅÅÅÅÅÅÅ
ÅÅ = ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ
-4
-4
1
ÅÅÅÅ
2
x
y = - ÅÅÅÅ54 + ÅÅÅÅ12 x
The slopes of the two lines are equal. So the lines are parallel.
HbL
The first line is 4 x + 2 y = 1 and the second line is 2 x - y = -1.
4x+2y= 1
-4 x
-4x
2y= 1-4x
2 x - y = -1
-2 x
-2x
-y = -1 - 2 x
2y
1-4 x
ÅÅÅÅ
ÅÅÅÅ = ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ
2
2
-y
-1-2 x
ÅÅÅÅ
ÅÅÅÅ = ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅÅ
-1
-1
y = ÅÅÅÅ12 - 2 x
y=1+2x
The slopes of the two lines are not equal and they are not negative reciprocals. So the lines are neither parallel
nor perpendicular.
HcL
The first line is x - 3 y = 2 and the second line is -3 x - y = 9.
x-3y=2
-x
-x
-3 y = 2 - x
-3 x - y = 9
+3 x
+3x
-y = 9 + 3 x
-3 y
2-x
ÅÅÅÅÅÅÅÅ
ÅÅ = ÅÅÅÅ
ÅÅÅÅÅÅ
-3
-3
-y
9+3 x
ÅÅÅÅ
ÅÅÅÅ = ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ
-1
-1
y = - ÅÅÅÅ23 + ÅÅÅÅ13 x
y = -9 - 3 x
The slopes of the two lines are negative reciprocals. So the lines are perpendicular.
19.
Given the function f HxL = 2 x2 + 3, find the following:
HaL
f H0L
f H-2L
HbL
20.
f H0L = 2 H0L2 + 3
= 2 H0L + 3
=0+3
=3
f H-2L = 2 H-2L2 + 3
= 2 H4L + 3
=8+3
= 11
Given the following graph of a relation, answer the questions.
5
4
3
2
1
-5
-4
-3
-2
-1
-1
-2
-3
-4
-5
1
2
3
4
5
HaL
Find the domain and the range.
HbL
Determine whether the relation is a function (give reasons for your answer).
Domain: H-¶, ¶L, Range: @-4, ¶L
This relation is a function. There is nowhere that a person could draw a vertical line that hits the graph more than
once.
21.
Given the following graph of a relation, answer the questions.
2
1
1
2
3
4
5
-1
-2
HaL
Find the domain and the range.
HbL
Determine whether the relation is a function (give reasons for your answer).
Domain: @0, ¶L, Range: H-¶, ¶L
It is not a function because it does not pass the vertical line test. For example, draw a vertical line at x = 2 and
this line hits the graph of the relation twice.