Math Practice Problem Set - Evans School of Public Affairs

EVANS SCHOOL OF PUBLIC POLICY AND GOVERNANCE
Math practice problems for incoming MPA students
Summer 2015
These math problems are for incoming students to practice some of the algebraic
operations that will be required in PBAF 516. This document is not a comprehensive
math review for PBAF 516 or for the Evans MPA. Instead, the goal of this document is
to provide practice on topics that tend to cause difficulties at the start of fall term for
students who do not like math or who have not taken a math course in many years.
Students who are able to quickly and correctly solve all of these problems when they
begin the MPA will have an advantage on quizzes and exams.
Section 6, on natural logarithms and exponentials, is more relevant for PBAF 528
(spring term) than PBAF 516.
Thanks to Jon Losey (MPA ’16) for putting together these problems.
1
1. INVERTING EQUATIONS
Go to https://www.mathsisfun.com/sets/function-inverse.html for more examples.
EXAMPLE
What is the inverse of the equation 𝑦 = 2π‘₯ + 6?
𝟏
The solution is: 𝒙 = π’š βˆ’ πŸ‘
𝟐
The steps to get there: 𝑦 = 2π‘₯ + 6 àοƒ  𝑦 βˆ’ 6 = 2π‘₯ àοƒ 
!!!
!
𝟏
= π‘₯ àοƒ  𝟐 π’š βˆ’ πŸ‘ = 𝒙
PRACTICE PROBLEMS
Solve for the inverse of the following equations:
a) 𝑦 = π‘₯ + 4
e) 𝑃 = .25𝑄 βˆ’ 100
b) π‘₯ = 2𝑦 + 4
f)
c) 𝑦 = 3π‘₯ + 5
!
𝑦 = βˆ’ π‘₯ βˆ’ 100
!
g) 𝑃 = βˆ’.1𝑄 + 400
d) 𝑦 = 4π‘₯ βˆ’ 8
!
h) 𝑦 = 32 + π‘₯
!
2
2. SOLVING PAIRS OF SIMULTANEOUS EQUATIONS
Go to http://www.themathpage.com/alg/simultaneous-equations.htm for more examples.
EXAMPLE
Solve for x and y in the following equations by using either addition/subtraction or substitution:
π‘₯+𝑦=1
π‘₯ + 3𝑦 = 9
The solution by substitution is: π‘₯ = βˆ’3, 𝑦 = 4
The steps: π‘₯ = 1 βˆ’ 𝑦 àοƒ  1 βˆ’ 𝑦 + 3𝑦 = 9 àοƒ  2𝑦 = 8 àοƒ  π’š = πŸ’. Then substitute y into either equation:
π‘₯ + 4 = 1 àοƒ  𝒙 = βˆ’πŸ‘
PRACTICE PROBLEMS
Find the values of x and y that solve the following pairs of equations:
a) 3π‘₯ + 𝑦 = 13
π‘₯ + 6𝑦 = βˆ’ 7
e) 4π‘₯ + 𝑦 = 9
π‘₯βˆ’π‘¦=1
b) 𝑦 = π‘₯ + 4
π‘₯ = 2𝑦 + 4
f)
c) 5π‘Ž = 2𝑏 + 3
2π‘Ž βˆ’ 𝑏 = 0
g) 11π‘₯ + 6𝑦 = 79
11π‘₯ + 3𝑦 = 67
d) π‘₯ + 2𝑦 = 8
π‘₯ βˆ’ 2𝑦 = 4
h)
3
2π‘₯ + 3𝑦 = 28
π‘₯ + 𝑦 = 11
!
!
π‘₯ + 𝑦 = 8
!
2
3
π‘₯ + 𝑦 = 17
3
2
!
3. GRAPHING FROM AN EQUATION OR ITS INVERSE
Graph the following equations with y on the vertical axis and x on the horizontal axis.
EXAMPLE
Graph 𝑦 = 2π‘₯ + 6
Solution:
PRACTICE PROBLEMS
a) Graph 𝑦 = 3π‘₯
b) Graph 𝑦 = π‘₯ + 3
4
!
c) Graph 𝑦 = π‘₯ + 1
!
!
d) Graph 𝑦 = π‘₯ + 8
!
e) Graph π‘₯ = 𝑦 + 3
g) Graph 𝑦 = 5π‘₯ and 𝑦 = βˆ’5π‘₯ + 50
!
f) Graph π‘₯ = 𝑦 + 1
!
!
!
!
!
h) Graph π‘₯ = 𝑦 βˆ’ 7.5 and 𝑦 = βˆ’ π‘₯ + 25
5
4. RECOVERING THE EQUATION OF A LINE FROM A GRAPH
Find more examples at https://www.mathsisfun.com/equation_of_line.html.
EXAMPLE
Find the equation for the line, with y as the vertical axis variable and x as the horizontal axis variable.
𝟐
Solution: Intercept is 9 and slope is 2/3, therefore: π’š = 𝒙 + πŸ—
πŸ‘
PRACTICE PROBLEMS
a) Find the equation for the line.
b) Find the equation for the line.
6
c) Find the equation for the line.
d) Find the equation for the line.
e) Find the equation for the line.
f) Find the equation for the line.
g) Find the equation for the line.
h) Find the equation for the line.
i) Find the equation for the line.
7
5. CALCULATING AREAS UNDER LINES
The goal in this section is to calculate the area between the graph and the horizontal axis.
EXAMPLE
Solution: (Base X Height) / 2 àοƒ  (20 X 8) / 2 àοƒ  80
PRACTICE PROBLEMS
a) Find the area under the line.
b) Find the area under the line.
8
c) Find the area under the line.
d) Find the area under the line.
e) Find the area under the line.
f) Find the area under the line.
g) Find the area under the line.
h) Find the area under the line.
i) Find the area under the line.
9
6. NATURAL LOGS AND EXPONENTIAL FUNCTIONS
Here are some of the rules of natural logs: http://www.rapidtables.com/math/algebra/Ln.htm. Here are some
additional problems for natural logs: http://www.millersville.edu/~bikenaga/basic-algebra/log/log.pdf.
EXAMPLE
Write the equation ln 20.09 = 3 in its exponential form.
Solution: If ln x = k, then x = ek. Therefore, 20.09 = e3
PRACTICE PROBLEMS
a) Express in exponential form: ln 2.7183 = 1
b) Write the equation e2.7 β‰ˆ 14.88 in logarithmic form.
c) Express the equation e2 β‰ˆ 7.39 in logarithmic form.
d) Write as a single logarithm: ln 3 + ln 7
e) Write as a single logarithm: ln 6 – ln 2
f)
Expand the following expression: ln 12π‘₯ !
g) Expand the following expression: ln
!! !
!!
h) Solve for x in the equation: ln x = 24
i)
Solve for x in the equation: e4x+2 = 50
j)
Solve for z in the equation: 1 + 2e1-3z = 15
10