Math Refresher, Unit 3:
Systems of Equations, Inequalities
Junesoo Lee PhD Candidate of Public Administration Instructor for Statistics (PUB316 and PAD505) [email protected]
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RPAD Welcome Week August 18-24, 2012
Agenda
Overview
Chapter 25: Functions
Chapter 26: Solving Systems of Equations
Chapter 28: Solving Inequalities—First Degree
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Book chapters from Forgotten Algebra (Barron’s)
On order at Mary Jane Books (see ad in Welcome Week booklet)
Overview
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• Why math in MPA?
o Policy and Management are basically numerical data driven
o Must in RPAD 501 (budgets), 503 (economics), 504 (data),
and 505 (statistics)
• You already know math well, but only forgot it
o The only thing we should blame OBLIVION on is TIME
• Let’s refresh what you must have known a million
years before
Agenda
Overview
Chapter 25: Functions
Chapter 26: Solving Systems of Equations
Chapter 28: Solving Inequalities—First Degree
4
Book chapters from Forgotten Algebra (Barron’s)
On order at Mary Jane Books (see ad in Welcome Week booklet)
Functions
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Terminology
Takes a number
and adds 3
X Domain Input Independent variable
f (or g, or h as
name of function)
Rule X + 3
Y Range Output Dependent variable
A function is a rule that assigns to each element in the domain one and only one element in the range. (Unless specified, the domain of a function is the set of all real numbers.)
𝑦 = 𝑓 𝑥 = 𝑥 + 3 𝑤𝑖𝑡ℎ 𝐷 = {5, 7, 8, 9}
"𝑦 𝑒𝑞𝑢𝑎𝑙𝑠 𝑓 𝑜𝑓 𝑥 𝑒𝑞𝑢𝑎𝑙𝑠 𝑥 𝑝𝑙𝑢𝑠 3, 𝑤𝑖𝑡ℎ 𝑑𝑜𝑚𝑎𝑖𝑛 𝐷 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 𝑡ℎ𝑒 𝑠𝑒𝑡 𝑜𝑓 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 5, 7, 8, 𝑎𝑛𝑑 9. "
Functions
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Examples
• Given this function, find each of the following:
𝑓 𝑥 = 𝑥 + 3 (𝑖. 𝑒. , 𝑦 = 𝑥 + 3)
o 𝑓(7)
o 𝑓(9)
o 𝑓−1(11)
= 7 + 3 = 10
= 9 + 3 = 12
11 = 𝑥 + 3 "𝑓 𝑖𝑛𝑣𝑒𝑟𝑠𝑒 𝑜𝑓 11"
"𝑖𝑓 𝑡ℎ𝑒 𝑦 𝑣𝑎𝑙𝑢𝑒 𝑖𝑠 11, 𝑤ℎ𝑎𝑡 𝑖𝑠 𝑥? "
𝑥 = 11 − 3 = 8
"𝑓 𝑜𝑓 7"
"𝑓 𝑜𝑓 9"
Functions
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Exercises
• Given this function, find each of the following:
𝑔 𝑥 = 1 + 𝑥2 (𝑖. 𝑒. , 𝑦 = 1 + 𝑥2)
o 𝑔(3)
o 𝑔(−2)
o 𝑔−1(10)
= 1 + 32 = 1 + 9 = 10
= 1 + −2 2 = 1 + 4 = 5
10 = 1 + 𝑥2 𝑥2 = 9
𝑥2 = 9 𝑥 = ±3
"𝑖𝑓 𝑡ℎ𝑒 𝑦 𝑣𝑎𝑙𝑢𝑒 𝑖𝑠 10, 𝑤ℎ𝑎𝑡 𝑖𝑠 𝑥? "
Agenda
Overview
Chapter 25: Functions
Chapter 26: Solving Systems of Equations
Chapter 28: Solving Inequalities—First Degree
8
Book chapters from Forgotten Algebra (Barron’s)
On order at Mary Jane Books (see ad in Welcome Week booklet)
Solving Systems of Equations
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Example
• Solve (i.e., find solutions): 𝑥 + 𝑦 = 8
o 𝑥 = 1 𝑎𝑛𝑑 𝑦 = 7. 𝑂𝑟 𝑠𝑖𝑚𝑝𝑙𝑦 1, 7
o 𝑥 = 2.5 𝑎𝑛𝑑 𝑦 = 5.5. 𝑂𝑟 2.5, 5.5
o 𝑥 = −3 𝑎𝑛𝑑 𝑦 = 11. 𝑂𝑟 (−3, 11)
o We can find infinitely many solutions to this equation.
An equation that has the form 𝑎𝑥 + 𝑏𝑦 = 𝑐, with 𝑎, 𝑏, and 𝑐 being real numbers, 𝑎 and 𝑏 not both zero, is a linear equation in two variables. The solutions to a system of equations are the pairs of values of 𝑥 and 𝑦 that satisfy all the equations in the system.
Solving Systems of Equations
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Example
• Solve:
o The solution to this system is 𝑥 = 10 𝑎𝑛𝑑 𝑦 = 4, or simply (10, 4).
o Prove it by plugging these values in the system.
A system of equations means that there is more than one equation related to one another.
2𝑥 + 𝑦 = 24 𝑥 − 𝑦 = 6
How can we find this solution? There are numerous ways to do it. But we will cover only two methods—Elimination by addition (or by substitution.)
2(10) + 4 = 24 10 − 4 = 6
Solving Systems of Equations
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Example
• Solve:
1. Write them in standard form.
2. Multiply (the second equation by -2 so that the y-coefficients are the negatives of one another.)
3. Add.
4. Solve.
Five steps of elimination by addition: 1. Write the equations in standard
form like 𝑎𝑥 + 𝑏𝑦 = 𝑐. 2. Multiply (if necessary) the
equations by constants so that the coefficients of the 𝑥 or the 𝑦 variable are the negatives of one another.
3. Add the equations from step 1. 4. Solve the equations from step 2. 5. Substitute the answer from step
3 back into one of the original equations, and solve for the second variable.
3𝑥 + 2𝑦 = 12 𝑦 = 2𝑥 − 1
3𝑥 + 2𝑦 = 12 −2𝑥 + 𝑦 = −1
3𝑥 + 2𝑦 = 12 −2 −2𝑥 + 𝑦 = −2(−1)
3𝑥 + 2𝑦 = 12 4𝑥 − 2𝑦 = 2
7𝑥 = 14 𝑥 = 2
5. Substitute back into an original equation. 𝐼𝑓 𝑥 = 2 𝑎𝑛𝑑 𝑦 = 2𝑥 − 1,
𝑡ℎ𝑒𝑛 𝑦 = 2 2 − 1, 𝑦 = 3.
Solution to system is 𝒙 = 𝟐 𝒂𝒏𝒅 𝒚 = 𝟑.
Solving Systems of Equations
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Example (continued)
• System:
• From an algebraic point of
view, 𝑥 = 2 𝑎𝑛𝑑 𝑦 = 3 is the
solution to this system.
• From a geometric point of
view, (2, 3) is the point of
intersection for two lines
whose equations are given
above.
3𝑥 + 2𝑦 = 12 𝑦 = 2𝑥 − 1
3𝑥 + 2𝑦 = 12 𝑦 = 2𝑥 − 1
(2, 3)
How to draw each line easily? Find x-intercept and y-intercept by plugging zero in x or y. And connect those intercepts.
(0, 6)
(4, 0) (1
2, 0)
(0, -1)
Solving Systems of Equations
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Exercise
• Solve:
1. Write them in standard form.
2. Multiply (the second equation by 1
5
so that the y-coefficients are the negatives of one another.)
3. Add.
4. Solve.
Five steps of elimination by addition: 1. Write the equations in standard
form like 𝑎𝑥 + 𝑏𝑦 = 𝑐. 2. Multiply (if necessary) the
equations by constants so that the coefficients of the 𝑥 or the 𝑦 variable are the negatives of one another.
3. Add the equations from step 1. 4. Solve the equations from step 2. 5. Substitute the answer from step
3 back into one of the original equations, and solve for the second variable.
3𝑥 − 𝑦 = −7 5𝑦 + 5 = −5𝑥
3𝑥 − 𝑦 = −7 5𝑥 + 5𝑦 = −5
3𝑥 − 𝑦 = −7 5𝑥
5+
5𝑦
5=
−5
5
3𝑥 − 𝑦 = −7 𝑥 + 𝑦 = −1
4𝑥 = −8 𝑥 = −2
5. Substitute back into an original equation. 𝐼𝑓 𝑥 = −2 𝑎𝑛𝑑 3𝑥 − 𝑦 = −7,
𝑡ℎ𝑒𝑛 3 −2 − 𝑦 = −7, 𝑦 = 1.
Solution to system is 𝒙 = −𝟐 𝒂𝒏𝒅 𝒚 = 𝟏.
Solving Systems of Equations
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Exercise (continued)
• System:
• Given the algebraic solution to
system (𝑥 = −2 𝑎𝑛𝑑 𝑦 = 1),
show and verify that the
solution point (-2, 1) also
makes sense from a geometric
point of view.
5𝑦 + 5 = −5𝑥 3𝑥 − 𝑦 = −7
(-2, 1)
3𝑥 − 𝑦 = −7 5𝑦 + 5 = −5𝑥
(0, 7)
(−7
3, 0) (-1, 0)
(0, -1)
How to draw each line easily? Find x-intercept and y-intercept by plugging zero in x or y. And connect those intercepts.
Solving Systems of Equations
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Example
• Solve:
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.
𝑦 + 8 = 2𝑥 3𝑥 + 2𝑦 = 12
𝑦 = 2𝑥 − 8
3𝑥 + 2𝑦 = 12
3𝑥 + 2(2𝑥 − 8) = 12
3𝑥 + 4𝑥 − 16 = 12 7𝑥 = 28
𝑥 = 4
𝐼𝑓 𝑥 = 4 𝑎𝑛𝑑 𝑦 + 8 = 2(4),
𝑡ℎ𝑒𝑛 𝑦 = 2 4 − 8, 𝑦 = 0.
Solution to system is 𝒙 = 𝟒 𝒂𝒏𝒅 𝒚 = 𝟎.
Solving Systems of Equations
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Exercise
• Solve:
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.
𝑦 − 1 = 3𝑥 3𝑥 + 4𝑦 = −26
𝑦 = 3𝑥 + 1
3𝑥 + 4𝑦 = −26
3𝑥 + 4 3𝑥 + 1 = −26
3𝑥 + 12𝑥 + 4 = −26 15𝑥 = −30
𝑥 = −2
𝐼𝑓 𝑥 = −2 𝑎𝑛𝑑 𝑦 − 1 = 3(−2),
𝑡ℎ𝑒𝑛 𝑦 = −6 + 1, 𝑦 = −5.
Solution to system is 𝒙 = −𝟐 𝒂𝒏𝒅 𝒚 = −𝟓.
Solving Systems of Equations
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Example
• Solve by using elimination by addition:
1. Write them in standard form.
2. Multiply (the second equation by -1 so that the y-coefficients are the negatives of one another.)
3. Add and solve.
The equations are said to be inconsistent when there is no solution to the system. Geometrically, two lines are parallel so there is no point of intersection for the two lines.
3𝑥 + 4𝑦 = 2 4𝑦 = 8 − 3𝑥
3𝑥 + 4𝑦 = 2 3𝑥 + 4𝑦 = 8
3𝑥 + 4𝑦 = 2 −1 3𝑥 + 4𝑦 = −1(8)
3𝑥 + 4𝑦 = 2 −3𝑥 − 4𝑦 = −8
0 = −6
False statement. There is no solution to the system.
Solving Systems of Equations
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Example (continued)
• System:
• From an algebraic point of view,
there is no solution to this
system.
• From a geometric point of view,
two lines are parallel so there is
no point of intersection for the
two lines.
(0, 2)
3𝑥 + 4𝑦 = 2 4𝑦 = 8 − 3𝑥
3𝑥 + 4𝑦 = 2
4𝑦 = 8 − 3𝑥
(8
3, 0) (0,
1
2) (
2
3, 0)
Solving Systems of Equations
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Example
• Solve by using elimination by addition:
1. Write them in standard form.
2. Multiply (the first equation by -2 so that the y-coefficients are the negatives of one another.)
3. Add and solve.
The equations are said to be dependent when there are infinitely many solutions to the system. Geometrically, two lines coincide so there are infinitely many points of intersection for the two lines.
3𝑥 − 𝑦 = 5 6𝑥 − 2𝑦 − 10 = 0
3𝑥 − 𝑦 = 5 6𝑥 − 2𝑦 = 10
−2 3𝑥 − 𝑦 = −2(5) 6𝑥 − 2𝑦 = 10
−6𝑥 + 2𝑦 = −10 6𝑥 − 2𝑦 = 10
0 = 0
True statement. All values of x and y satisfy this equation.
Solving Systems of Equations
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Example (continued)
• System:
• From an algebraic point of view,
there are infinitely many
solutions to the system.
• From a geometric point of view,
two lines coincide so all points
of the two lines intersect.
(5
3, 0)
3𝑥 − 𝑦 = 5 6𝑥 − 2𝑦 − 10 = 0 3𝑥 − 𝑦 = 5
6𝑥 − 2𝑦 − 10 = 0
(0, -5)
Solving Systems of Equations
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Exercise
• Solve, and figure out if two equations are inconsistent (i.e., no
solution) or dependent (i.e., infinite solutions).
1. Write them in standard form.
2. Multiply (the first equation by -2 so that the y-coefficients are the negatives of one another.)
3. Add and solve.
5𝑥 − 𝑦 = 4 2𝑦 = 10𝑥 − 6
5𝑥 − 𝑦 = 4 10𝑥 − 2𝑦 = 6
−2 5𝑥 − 𝑦 = −2(4) 10𝑥 − 2𝑦 = 6
0 = −2
False statement. There is no solution to the system.
−10𝑥 + 2𝑦 = −8 10𝑥 − 2𝑦 = 6
Solving Systems of Equations
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Exercise (continued)
• System:
• Given the algebraic solution to
system (no solution), show and
verify that this result also makes
sense from a geometric point of
view.
(4
5, 0)
5𝑥 − 𝑦 = 4 2𝑦 = 10𝑥 − 6
(0, -4)
5𝑥 − 𝑦 = 4
(3
5, 0)
(0, -3)
2𝑦 = 10𝑥 − 6
Agenda
Overview
Chapter 25: Functions
Chapter 26: Solving Systems of Equations
Chapter 28: Solving Inequalities—First Degree
23
Book chapters from Forgotten Algebra (Barron’s)
On order at Mary Jane Books (see ad in Welcome Week booklet)
Solving Inequalities—First Degree
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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.”
-2 3 <
Number line
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.
Solving Inequalities—First Degree
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Examples
• Solve: 𝑥 + 5 < 7
then 𝑥 < 7 − 5
and 𝑥 < 2.
• Solve: 1 − 𝑥 ≤ −2
then 1 + 2 ≤ 𝑥
and 3 ≤ 𝑥.
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.
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.
• Graphically represent the solutions
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 the answer.
Solving Inequalities—First Degree
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Examples
• 6 < 15, divided by 3
then 6
3<
15
3
and 2 < 5.
•1
4< 12, multiplied by 4
then 41
4< 4 12
and 1 < 48.
Rule 2: Reverse the direction of an inequality symbol whenever an inequality is multiplied or divided by the same negative number.
• 10 < 15, divided by -5
then 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).
Solving Inequalities—First Degree
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Example
• Solve: 4𝑥 − 3 > 6𝑥 + 1
o 4𝑥 − 3 > 6𝑥 + 1
4𝑥 − 6𝑥 > 1 + 3
−2𝑥 > 4
−2𝑥
−2<
4
−2
𝑥 < −2
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
Solving Inequalities—First Degree
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Exercise
• Solve: 4 𝑥 − 3 ≥ 8𝑥 − 4
o 4𝑥 − 12 ≥ 8𝑥 − 4
4𝑥 − 8𝑥 ≥ −4 + 12
−4𝑥 ≥ 8
−4𝑥
−4≤
8
−4
𝑥 ≤ −2
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
We are ready to start MPA!
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