Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables

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Systems in Two VariablesGraphing

SubstitutionElimination by Addition

Applications

Math 1300 Finite MathematicsSection 4.1 Review: Systems of Linear Equations in Two

Variables

Jason Aubrey

Department of MathematicsUniversity of Missouri

Jason Aubrey Math 1300 Finite Mathematics

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Applications

Definition (Systems of Equations in Two Variables)Given the linear system

ax + by = hcx + dy = k

A pair of numbers x = x0, y = y0 [also written as an orderedpair (x0, y0) is said to be a solution of the system if eachequation is satisfied by the pair. The set of all such orderedpairs is called the solution set for the system. To solve asystem is to find its solution set.

We will consider three methods for solving such systems:graphing, substitution, and elimination by addition.


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Applications

Definition (Systems of Equations in Two Variables)Given the linear system


A pair of numbers x = x0, y = y0 [also written as an orderedpair (x0, y0) is said to be a solution of the system if eachequation is satisfied by the pair. The set of all such orderedpairs is called the solution set for the system. To solve asystem is to find its solution set.

We will consider three methods for solving such systems:graphing, substitution, and elimination by addition.


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Applications

Example: Solve the linear system by graphing:

3x − y = 2x + 2y = 10

−4 −2 2 4 6 8 10−2

2

4

6

8

10

(2, 4)


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Applications


3x − y = 2x + 2y = 10

−4 −2 2 4 6 8 10−2

2

4

6

8

10

(2, 4)


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Applications


3x − y = 2x + 2y = 10

−4 −2 2 4 6 8 10−2

2

4

6

8

10

(2, 4)


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Applications


3x − y = 2x + 2y = 10

−4 −2 2 4 6 8 10−2

2

4

6

8

10

(2, 4)


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Applications

Definition (Systems of Linear Equations: Basic Terms)A system of linear equations is consistent if it has one or moresolutions and inconsistent if no solutions exist. Furthermore, aconsistent system is said to be independent if it has exactlyone solution and dependent if it has more than one solution.Two systems of equations are equivalent if they have the samesolution set.


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Applications

Theorem (Possible Solutions to a Linear System)The linear system


Must haveExactly one solution (consistent and independent), orNo solution (inconsistent), orInfinitely many solutions (consistent and dependent).

There are no other possibilities.


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Applications



Must haveExactly one solution (consistent and independent), or

No solution (inconsistent), orInfinitely many solutions (consistent and dependent).



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Applications



Must haveExactly one solution (consistent and independent), orNo solution (inconsistent), or

Infinitely many solutions (consistent and dependent).There are no other possibilities.


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Applications






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Applications






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Applications

Example: Graph the equations and find the coordinates of anypoints where two or more lines intersect. Discuss the nature ofthe solution set.

x − 2y = −62x + y = 8x + 2y = −2


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Applications

−4 −2 2 4

−2

2

4

6

0

h(2, 4)

(−2, 2)

(143 ,−4

3)

x − 2y = −62x + y = 8x + 2y = −2

No point lies on allthree lines. So, no so-lution to this system.


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Applications

−4 −2 2 4

−2

2

4

6

0

h(2, 4)

(−2, 2)

(143 ,−4

3)

x − 2y = −6

2x + y = 8x + 2y = −2



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Applications

−4 −2 2 4

−2

2

4

6

0

h(2, 4)

(−2, 2)

(143 ,−4

3)

x − 2y = −62x + y = 8

x + 2y = −2



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Applications

−4 −2 2 4

−2

2

4

6

0

h

(2, 4)

(−2, 2)

(143 ,−4

3)

x − 2y = −62x + y = 8x + 2y = −2



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Applications

−4 −2 2 4

−2

2

4

6

0

h(2, 4)

(−2, 2)

(143 ,−4

3)

x − 2y = −62x + y = 8x + 2y = −2



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Applications

−4 −2 2 4

−2

2

4

6

0

h(2, 4)

(−2, 2)

(143 ,−4

3)

x − 2y = −62x + y = 8x + 2y = −2



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Applications

−4 −2 2 4

−2

2

4

6

0

h(2, 4)

(−2, 2)

(143 ,−4

3)

x − 2y = −62x + y = 8x + 2y = −2



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Applications

Example: Solve the linear system by substitution:

2x + y = 6x − y = −3

2x + y = 6y = 6− 2x now substitute:

x − y = −3x − (6− 2x) = −3

3x − 6 = −3x = 1

Substituting we obtain y = 4. So, the solution is the point (1, 4).


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Applications


2x + y = 6x − y = −3

2x + y = 6

y = 6− 2x now substitute:x − y = −3

x − (6− 2x) = −33x − 6 = −3

x = 1



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Applications


2x + y = 6x − y = −3


x − y = −3x − (6− 2x) = −3

3x − 6 = −3x = 1



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Applications


2x + y = 6x − y = −3


x − y = −3

x − (6− 2x) = −33x − 6 = −3

x = 1



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Applications


2x + y = 6x − y = −3


x − y = −3x − (6− 2x) = −3

3x − 6 = −3x = 1



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Applications


2x + y = 6x − y = −3


x − y = −3x − (6− 2x) = −3

3x − 6 = −3

x = 1



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Applications


2x + y = 6x − y = −3


x − y = −3x − (6− 2x) = −3

3x − 6 = −3x = 1



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Applications


2x + y = 6x − y = −3


x − y = −3x − (6− 2x) = −3

3x − 6 = −3x = 1



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Applications

Graphing and substitution work well for systems involvingtwo variables.

However, neither is easily extended to larger systems.Elimination by addition is the most important method ofsolution.It readily generalizes to larger systems and forms the basisfor computer-based solution methods.


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Applications

Graphing and substitution work well for systems involvingtwo variables.However, neither is easily extended to larger systems.

Elimination by addition is the most important method ofsolution.It readily generalizes to larger systems and forms the basisfor computer-based solution methods.


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Applications

Graphing and substitution work well for systems involvingtwo variables.However, neither is easily extended to larger systems.Elimination by addition is the most important method ofsolution.

It readily generalizes to larger systems and forms the basisfor computer-based solution methods.


university-logo



Applications

Graphing and substitution work well for systems involvingtwo variables.However, neither is easily extended to larger systems.Elimination by addition is the most important method ofsolution.It readily generalizes to larger systems and forms the basisfor computer-based solution methods.


university-logo



Applications

Theorem (Operations that Produce Equivalent Systems)A system of linear equations is transformed into an equivalentsystem if

(A) Two equations are interchanged.(B) An equation is multiplied by a nonzero constant.(C) A constant multiple of one equation is added to another

equation.


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Applications

Theorem (Operations that Produce Equivalent Systems)A system of linear equations is transformed into an equivalentsystem if(A) Two equations are interchanged.

(B) An equation is multiplied by a nonzero constant.(C) A constant multiple of one equation is added to another

equation.


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Applications

Theorem (Operations that Produce Equivalent Systems)A system of linear equations is transformed into an equivalentsystem if(A) Two equations are interchanged.(B) An equation is multiplied by a nonzero constant.

(C) A constant multiple of one equation is added to anotherequation.


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Applications

Theorem (Operations that Produce Equivalent Systems)A system of linear equations is transformed into an equivalentsystem if(A) Two equations are interchanged.(B) An equation is multiplied by a nonzero constant.(C) A constant multiple of one equation is added to another

equation.


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Applications

Example: Solve the linear system using elimination by addition

3u − 2v = 127u + 2v = 8

3u − 2v = 12+7u + 2v = 810u + 0 = 20

u = 2

Substituting, we obtain v = −3. So the solution is (2,-3)


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Applications


3u − 2v = 127u + 2v = 8

3u − 2v = 12

+7u + 2v = 810u + 0 = 20

u = 2



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Applications


3u − 2v = 127u + 2v = 8

3u − 2v = 12+7u + 2v = 8

10u + 0 = 20u = 2



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Applications


3u − 2v = 127u + 2v = 8

3u − 2v = 12+7u + 2v = 810u + 0 = 20

u = 2



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Applications


3u − 2v = 127u + 2v = 8

3u − 2v = 12+7u + 2v = 810u + 0 = 20

u = 2



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Applications


3u − 2v = 127u + 2v = 8

3u − 2v = 12+7u + 2v = 810u + 0 = 20

u = 2



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Applications

Example: Solve using elimination by addition.

3x − 2y = 82x + 5y = −1

We multiply the first equation by 5 and the bottom equation by 2and then add.

15x − 10y = 404x + 10y = −2

19x = 38

Now we multiply both sides of this last equation by 119 to obtain

x = 2. Then we substitute back into either of the two originalequations to obtain y = −1.


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Applications


3x − 2y = 82x + 5y = −1


15x − 10y = 404x + 10y = −2

19x = 38




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Applications


3x − 2y = 82x + 5y = −1


15x − 10y = 40

4x + 10y = −219x = 38




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Applications


3x − 2y = 82x + 5y = −1


15x − 10y = 404x + 10y = −2

19x = 38




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Applications


3x − 2y = 82x + 5y = −1


15x − 10y = 404x + 10y = −2

19x = 38




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Applications


3x − 2y = 82x + 5y = −1


15x − 10y = 404x + 10y = −2

19x = 38


x = 2.

Then we substitute back into either of the two originalequations to obtain y = −1.


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Applications


3x − 2y = 82x + 5y = −1


15x − 10y = 404x + 10y = −2

19x = 38




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Applications

Example: At $4.80 per bushel, the annual supply for soybeansin the Midwest is 1.9 billion bushels and the annual demand is2.0 billion bushels. When the price increases to $5.10 perbushel, the annual supply increases to 2.1 billion bushels andthe annual demand decreases to 1.8 billion bushels. Assumethat the supply and demand equations are linear.


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Applications

(a) Find the supply equation.

We wish to find an equation of the form p = mq + b where prepresents unit price and q represents quantity demanded.

We have two points (q, p) on the graph of the supply equation:(1.9, 4.80) and (2.1, 5.10).

The slope of this line is:

m =5.10− 4.80

2.1− 1.9=

0.30.2

=32


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Applications





m =5.10− 4.80

2.1− 1.9=

0.30.2

=32


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Applications





m =5.10− 4.80

2.1− 1.9=

0.30.2

=32


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Applications





m =5.10− 4.80

2.1− 1.9=

0.30.2

=32


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Applications

Therefore the equation of this line is

p − 4.8 =32(q − 1.9)

p =32

q − 32(1.9) + 4.8

p =32

q + 1.95


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Applications


p − 4.8 =32(q − 1.9)

p =32

q − 32(1.9) + 4.8

p =32

q + 1.95


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Applications


p − 4.8 =32(q − 1.9)

p =32

q − 32(1.9) + 4.8

p =32

q + 1.95


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Applications


p − 4.8 =32(q − 1.9)

p =32

q − 32(1.9) + 4.8

p =32

q + 1.95


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Applications

(b) Find the demand equation.

Again, we want an equation of the form p = mq + b. We havetwo points on the graph of the demand equation: (2.0, 4.80)and (1.8, 5.10).


m =5.10− 4.80

1.8− 2.0= −0.3

0.2= −3

2Therefore the equation of this line is

p − 4.80 = −32(q − 2.0)

p = −32

q +32(2.0) + 4.80

p = −32

q + 7.80


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Applications




m =5.10− 4.80

1.8− 2.0= −0.3

0.2= −3


p − 4.80 = −32(q − 2.0)

p = −32

q +32(2.0) + 4.80

p = −32

q + 7.80


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Applications




m =5.10− 4.80

1.8− 2.0= −0.3

0.2= −3

2


p − 4.80 = −32(q − 2.0)

p = −32

q +32(2.0) + 4.80

p = −32

q + 7.80


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Applications




m =5.10− 4.80

1.8− 2.0= −0.3

0.2= −3


p − 4.80 = −32(q − 2.0)

p = −32

q +32(2.0) + 4.80

p = −32

q + 7.80


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Applications




m =5.10− 4.80

1.8− 2.0= −0.3

0.2= −3


p − 4.80 = −32(q − 2.0)

p = −32

q +32(2.0) + 4.80

p = −32

q + 7.80


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Applications




m =5.10− 4.80

1.8− 2.0= −0.3

0.2= −3


p − 4.80 = −32(q − 2.0)

p = −32

q +32(2.0) + 4.80

p = −32

q + 7.80


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Applications

(c) Find the equilibrium price and quantity.

We set supply equal to demand and solve:

− 32

q + 7.80 =32

q + 1.95

3q = 5.85q = 1.95 billion bushels

Substituting, we find p = −32(1.95) + 7.80 = $4.88


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Applications



− 32

q + 7.80 =32

q + 1.95




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Applications



− 32

q + 7.80 =32

q + 1.95

3q = 5.85

q = 1.95 billion bushels



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Applications



− 32

q + 7.80 =32

q + 1.95




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Applications



− 32

q + 7.80 =32

q + 1.95




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Applications

(d) Graph the two equations in the same coordinate system andidentify the equilibrium point, supply curve, and demand curve.

2 4 6 8 10 12

2

4

6

0

Supply Curve

Demand Curve

f

g

(1.95, 4.88)


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Applications


2 4 6 8 10 12

2

4

6

0

Supply Curve

Demand Curve

f

g

(1.95, 4.88)


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Applications


2 4 6 8 10 12

2

4

6

0

Supply Curve

Demand Curve

f

g

(1.95, 4.88)


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Applications


2 4 6 8 10 12

2

4

6

0

Supply Curve

Demand Curve

f

g

(1.95, 4.88)


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Applications


2 4 6 8 10 12

2

4

6

0

Supply Curve

Demand Curve

f

g

(1.95, 4.88)


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Applications


2 4 6 8 10 12

2

4

6

0

Supply Curve

Demand Curve

f

g

(1.95, 4.88)


Education

Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables