RPAD Welcome Week
Math Refresher, III:
Function, Systems of Equations, Inequalities
Gang Chen
Assistant Professor
gchen3@albany.edu
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• Gang Chen, assistant professor
• Financial management, budget analysis,
pension management
• PAD 501 financial management for public and
nonprofit organizations
• PAD 642 public budgeting
• PAD 505 Data, Models, and Decisions
Concerns
• Are you good at Math?
• Quantitative courses in MPA program:
– 501, 503, 504, 505
– Elective courses
• Why learning quantitative skills?
– Be competitive in job market
– Consumer of quantitative reports
Path to success
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Be confident!
Find good resources
Basic rules
Follow step-by-step guide
Practices
Office hours!!!
Functions
Exercises
• Given this function, find each of the following:
𝑓 𝑥 = 1 + 2𝑥 2 (𝑖. 𝑒. , 𝑦 = 1 + 𝑥 2 )
o 𝑓(3) =
o 𝑓(−2) =
o 𝑓 −1 (10)
"𝑖𝑓 𝑡ℎ𝑒 𝑦 𝑣𝑎𝑙𝑢𝑒 𝑖𝑠 10, 𝑤ℎ𝑎𝑡 𝑖𝑠 𝑥? "
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Solving Systems of Equations
Example
Five steps of elimination by addition:
1. Write the equations in standard
3𝑥 + 2𝑦 = 12
• Solve:
form like 𝑎𝑥 + 𝑏𝑦 = 𝑐.
𝑦 = 2𝑥 − 1
2. Multiply (if necessary) the
1. Write them in standard form.
equations by constants so that
the coefficients of the 𝑥 or the 𝑦
variable are the negatives of one
2. Multiply (the second equation by -2
another.
so that the y-coefficients are the
3. Add the equations from step 1.
4. Solve the equations from step 2.
negatives of one another.)
5. Substitute the answer from step
3 back into one of the original
equations, and solve for the
second variable.
3.
Add.
4.
Solve.
5. Substitute back into an original equation.
Solution to system is 𝒙 = 𝟐 𝒂𝒏𝒅 𝒚 = 𝟑.
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Solving Systems of Equations
Example (continued)
3𝑥 + 2𝑦 = 12
• System:
𝑦 = 2𝑥 − 1
3𝑥 + 2𝑦 = 12
𝑦 = 2𝑥 − 1
(0, 6)
• From an algebraic point of
(2, 3)
view, 𝑥 = 2 𝑎𝑛𝑑 𝑦 = 3 is the
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solution to this system.
(2, 0)
• From a geometric point of
(0, -1)
(4, 0)
view, (2, 3) is the point of
intersection for two lines
whose equations are given
above.
How to draw each line easily?
Find x-intercept and y-intercept by plugging
zero in x or y. And connect those intercepts.
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Solving Systems of Equations
Example
• Solve:
𝑦 + 8 = 2𝑥
3𝑥 + 2𝑦 = 12
1.
Find (or make) any variable
having a coefficient of 1, and
isolate it.
2.
Use the isolated variable with
a coefficient of 1 to replace
that in the other equation.
3.
Finish the problem.
Three steps of elimination by substitution:
1. Find any variable with a coefficient of 1,
or make any variable so, and isolate it in
one equation like 𝑦 = 𝑎𝑥 + 𝑏, 𝑜𝑟 𝑥 =
𝑎𝑦 + 𝑏.
2. Use the equation having the variable
with a coefficient of 1 to replace that
variable in the other equation.
3. Finish the problem as before by
substituting back into an original
equation.
Solution to system is 𝒙 = 𝟒 𝒂𝒏𝒅 𝒚 = 𝟎.
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Solving Inequalities—First Degree
Examples for inequality signs
• 2 < 3 is read “2 is less than 3.”
• 5 > 1 is read “5 is greater than 1.”
• 𝑎 ≤ 4 is read “𝑎 is less than or equal to 4.”
• 𝑏 ≥ 7 is read “𝑏 is greater than or equal to 7.”
Both expressions −2 < 3 and 3 > −2 have the same meaning.
But −2 < 3 is a better way because it clearly visualizes the
direction of difference like the number line.
Number line
-2
<
3
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Solving Inequalities—First Degree
Examples
To solve a first-degree inequality, find
the values of 𝑥 that satisfy the inequality.
The basic strategy is the same as that
used to solve first-degree equations.
• Solve: 𝑥 + 5 < 7
then 𝑥 < 7 − 5
Rule 1: A term may be transposed from
one side of the inequality to the other by
changing its sign as it crosses the
inequality sign.
and 𝑥 < 2.
• Solve: 1 − 𝑥 ≤ −2
•
Graphically represent the solutions
then 1 + 2 ≤ 𝑥
and 3 ≤ 𝑥.
The heavy line indicates that all numbers to the left
of 2 (or to the right of 3) are part of the answer.
The open circle indicates that 2 is not part of the
answer. The closed circle indicates that 3 is a part
of
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the answer.
Solving Inequalities—First Degree
Examples
•
6 < 15, divided by 3
6
then
3
<
15
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Rule 2: Reverse the direction of an
inequality symbol whenever an
inequality is multiplied or divided by the
same negative number.
•
then
and 2 < 5.
•
1
4
< 12, multiplied by 4
then 4
1
4
10 < 15, divided by -5
< 4 12
and 1 < 48.
10
−5
>
15
−5
and −2 > −3. (Or −3 < −2).
•
−𝑥
2
≤ 8, multiplied by -2
then −2
−𝑥
2
≥ −2(8)
and 𝑥 ≥ −16. (Or −16 ≤ 𝑥).
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Solving Inequalities—First Degree
Exercise
• Solve: 4 𝑥 − 3 ≥ 8𝑥 − 4
Rule 1: A term may be transposed from
one side of the inequality to the other by
changing its sign as it crosses the
inequality sign.
Rule 2: Reverse the direction of an
inequality symbol whenever an
inequality is multiplied or divided by the
same negative number.
• Graphically represent the solution
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We are ready to start MPA!
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