Download Math 115: Elementary Algebra Instructor: Julio C. Herrera Exam 2

Math 115: Elementary Algebra
Instructor: Julio C. Herrera
Exam 2
Name:
March 9, 2015
Exam Score:
Instructions: This exam covers the material from chapter two. 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: Solve
1
z
=
for z; use the multiplication principle.
9
−5
Solution:
Start off by cross multiplying and then isolate z through division.
1
z
=
9
−5
⇒ −5 = 9z
5
⇒− =z
9
5
That is, z = − .
9
Problem 2: Clear fractions or decimals, solve, and check.
2
3
7
5
3
− 4x − =
8
8
8
Solution:
2
3
7
5
3
− 4x − =
8
8
8
⇒
14 8
5
3
− x− =
24 3
8
8
⇒
14 64
15
9
− x−
=
24 24
24
24
(use the LCD 24)
⇒
14 15
9
64
−
−
= x
24 24 24
24
(isolate terms with x’s)
⇒−
10
64
= x
24
24
(combine like terms)
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Math 115: Elementary Algebra
Instructor: Julio C. Herrera
10
24
10
⇒−
64
5
⇒−
32
⇒−
Thus, x = −
·
24
=x
64
Exam 2
March 9, 2015
(isolate x)
=x
(simplify)
=x
(reduce)
5
32
Problem 3: The formula A = 12 ah + 21 bh can be used to find the area A of a trapezoid
with bases a and b and height h. Solve for h. (Hint: First clear fractions.)
Solution:
A = 12 ah + 12 bh
⇒ A = 12 h(a + b)
(factor 12 h)
⇒ 2A = h(a + b)
(multiply both sides by 2)
⇒
2A
=h
(a + b)
Hence, h =
(divide both sides by (a + b))
2A
.
(a + b)
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Math 115: Elementary Algebra
Instructor: Julio C. Herrera
Exam 2
March 9, 2015
Problem 4: The second angle of an architect’s triangle is three times as large as the first.
The third angle is 30◦ more than the first. Find the measure of each angle.
Solution:
With time you’ll learn that these type of problems are among the easiest and most fun
to solve. Let’s start by naming our first, second, and third angle x, y, and z, respectively.
We have three angles and all of their measurements are given to us in terms of the first
angle. Start by translating the information from English to math.
Second Angle: The second angle of an architect’s triangle is three times as large as the first.
y = 3x
Third Angle: The third angle is 30◦ more than the first.
z = x + 30◦
We know that these angles have to sum up to 180◦ . That is, x + y + z = 180. Let’s plug
in the information we know into this equation.
x + 3x + x + 30 = 180
Solve for x, which you should know how to do. x = 30. Plug this value of x into y and z
and you get that y = 3x = 3(30) = 90 and z = x + 30 = 30 + 30 = 60. Hence, the measure
of the first, second, and third angle are 30, 90, and 60, respectively.
Problem 5: Solve using the addition and multiplication principles.
3(2y − 3) > 21
Solution:
3(2y − 3) > 21
⇒ 6y − 9 > 21
⇒ 6y > 30
⇒y>5
Hence, the solution set is {y|y > 5} or (5, ∞).
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