Math 115: Elementary Algebra
Instructor: Julio C. Herrera
Exam 3
February 2, 2016
Name:
Exam Score:
Instructions: This exam covers the material from chapter three and eight. Please read
each question carefully before you attempt to solve it. Remember that you have to show all
of your work clearly if you want to get credit. The exam is closed book and calculators are
not allowed. Good luck!
Problem 1: Plot the ordered pairs in the table on the rectangular coordinate system.
Solution:
y
9
8
7
x y
0 1
2 0
-2 0
0 -3
1 3
-2 4
-3 -4
-5 6
8 -6
6
5
4
3
2
1
-8 -7 -6 -5 -4 -3 -2 -1
-1
-2
-3
-4
-5
-6
-7
-8
-9
Page 1 of 6
1 2
3
4
5
6
7
8
9
x
Math 115: Elementary Algebra
Instructor: Julio C. Herrera
Exam 3
Problem 2: Use the linear function f (x) =
February 2, 2016
3x − 2
to complete the table of values.
5
Solution:
x
0
2
5
11
-3
-5
y
− 2/5
4/5
13/5
31/5
− 11/5
− 17/5
f (0) =
3(0) − 2
−2
=
5
5
f (2) =
4
3(2) − 2
=
5
5
f (5) =
3(5) − 2
13
=
5
5
f (11) =
31
3(11) − 2
=
5
5
f (−3) =
3(−3) − 2
−11
=
5
5
f (−5) =
−17
3(−5) − 2
=
5
5
Problem 3: Rewrite the linear equation 2x + 3(x + y) = x − y + 2 in the form y = mx + b.
Solution:
2x + 3(x + y) = x − y + 2
⇒ 2x + 3x + 3y = x − y + 2
⇒ 5x + 3y = x − y + 2
⇒ 5x + 4y = x + 2
⇒ 4y = −4x + 2
2
4
1
⇒ y = −x +
2
⇒ y = −x +
Page 2 of 6
Math 115: Elementary Algebra
Instructor: Julio C. Herrera
Exam 3
February 2, 2016
x 3y
= 3 and then
Problem 4: Calculate the x-intercept and y-intercept for the line +
3
2
use the intercepts to graph the line.
Solution:
Remember that the x-intercept is of the form (x, 0) and the y-intercept is of the form (0, y).
x-intercept: Plug y = 0 into the equation.
y
x 3(0)
+
=3
3
2
x
⇒ =3
3
⇒x=9
9
8
7
6
5
The x-intercept is (9, 0)
4
y-intercept: Plug x = 0 into the equation.
0 3y
+
=3
3
2
3y
=3
⇒
2
⇒ 3y = 6
6
⇒y=
3
⇒y=2
The y-intercept is (0, 2)
Page 3 of 6
3
2
1
-8 -7 -6 -5 -4 -3 -2 -1
-1
-2
-3
-4
-5
-6
-7
-8
-9
1 2
3
4
5
6
7
8
9
x
Math 115: Elementary Algebra
Instructor: Julio C. Herrera
Exam 3
February 2, 2016
Problem 5: Write in the slope-intercept form the equation of the line passing through the
points (−2, 1) and (3, 2).
Solution:
We need to find the line y = mx + b that passes through the two points (−2, 1) and (3, 2).
First, let’s find the slope:
m=
2−1
1
y2 − y1
=
=
x2 − x1
3 − (−2)
5
Now that we have the slope of the line we can use any of the two points on the line and the
slope-point formula y − y1 = m(x − x1 ) to find our line.
Let’s use the point (3, 2):
1
y − 2 = (x − 3)
5
Solving for y yields:
1
y − 2 = (x − 3)
5
1
⇒ y = (x − 3) + 2
5
3
1
⇒y = x− +2
5
5
1
3 10
⇒y = x− +
5
5
5
1
7
⇒y = x+
5
5
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Math 115: Elementary Algebra
Instructor: Julio C. Herrera
Exam 3
February 2, 2016
Problem 6: Solve the following system of linear equations graphically.
2x − 3y = 3
4x − 3y = 9
Solution:
The first thing to be done is to to rewrite each equation the in slope intercept form
y = mx + b.
2
y = x−1
3
4
y = x−3
3
Graphing lines in these form should be super easy for you at this point. You can see these
lines in the following graph.
Notice that (3, 1) is the only point that
satisfies both equations; it is the only
point that you can plug its x-coordinate
into both equations and get the same ycoordinate in both equations.
Page 5 of 6
Math 115: Elementary Algebra
Instructor: Julio C. Herrera
Exam 3
February 2, 2016
Problem 7: Elyse jogs and walks to school each day. She averages 4 km/h walking and 8
km/h jogging. From home to school is 6 km and Elyse makes the trip in 1 hr. How far
does she jog in a trip?
Solution:
We will use the distance formula D = R · T to solve this problem. Let w be the walking
distance and j be the jogging distance. Now let’s look at what we know regarding walking
and jogging.
Walking: We know that w = r1 t1 . Notice that we are only given r1 = 4km/h. Hence,
w = 4t1
Jogging: We know that j = r2 t2 . We are given that r2 = 8km/h. Also notice that if Elyse
took one hour for the whole trip and she walked for t1 , then she must have jogged for the
remaining time of 1 hour minus t1 . We can write this mathematically as t2 = 1 − t1 . Hence,
j = 8(1 − t1 ) = 8 − 8t1 .
We know that the entire distance D = 6 has to equal the jogging and walking distance:
6 = D = w + j = 4t1 + 8 − 8t1 = 8 − 4t1
⇒ 6 = 8 − 4t1
⇒ 4t1 = 2
1
⇒ t1 =
2
Now that we know Elyse’s walking time we can find her walking and jogging distance.
1
Elyse walked for w = 4t1 = 4( ) = 2 km.
2
1
Elyse jogged for j = 8 − 8t1 = 8 − 8( ) = 4 km in a trip.
2
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