Math Refresher, Unit 3: Systems of Equations, Inequalities · PDF fileSystems of Equations, Inequalities Junesoo Lee ... Chapter 28: Solving Inequalities—First Degree 2 ... 3 +4

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




4



Functions

5

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

6

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

7

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




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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.


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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


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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.


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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)


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Exercise

• Solve:


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.


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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.


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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.


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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.


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Example

• Solve by using elimination by addition:


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.


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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)


19

Example

• Solve by using elimination by addition:


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.


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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)


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Exercise

• Solve, and figure out if two equations are inconsistent (i.e., no

solution) or dependent (i.e., infinite solutions).


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


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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




23



Solving Inequalities—First Degree

24

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.


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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.


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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).


27

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


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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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Math Refresher, Unit 3: Systems of Equations, Inequalities · PDF fileSystems of Equations, Inequalities Junesoo Lee ... Chapter 28: Solving Inequalities—First Degree 2 ... 3 +4