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Lecture over Section 4-1 of Barnett's "Finite Mathematics."
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university-logo
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
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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.
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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.
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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)
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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)
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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)
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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)
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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.
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
Applications
Theorem (Possible Solutions to a Linear System)The linear system
ax + by = hcx + dy = k
Must haveExactly one solution (consistent and independent), orNo solution (inconsistent), orInfinitely many solutions (consistent and dependent).
There are no other possibilities.
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
Applications
Theorem (Possible Solutions to a Linear System)The linear system
ax + by = hcx + dy = k
Must haveExactly one solution (consistent and independent), or
No solution (inconsistent), orInfinitely many solutions (consistent and dependent).
There are no other possibilities.
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
Applications
Theorem (Possible Solutions to a Linear System)The linear system
ax + by = hcx + dy = k
Must haveExactly one solution (consistent and independent), orNo solution (inconsistent), or
Infinitely many solutions (consistent and dependent).There are no other possibilities.
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
Applications
Theorem (Possible Solutions to a Linear System)The linear system
ax + by = hcx + dy = k
Must haveExactly one solution (consistent and independent), orNo solution (inconsistent), orInfinitely many solutions (consistent and dependent).
There are no other possibilities.
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
Applications
Theorem (Possible Solutions to a Linear System)The linear system
ax + by = hcx + dy = k
Must haveExactly one solution (consistent and independent), orNo solution (inconsistent), orInfinitely many solutions (consistent and dependent).
There are no other possibilities.
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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.
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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
No point lies on allthree lines. So, no so-lution to this system.
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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
No point lies on allthree lines. So, no so-lution to this system.
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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.
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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.
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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.
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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.
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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).
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
Applications
Example: Solve the linear system by substitution:
2x + y = 6x − y = −3
2x + y = 6
y = 6− 2x now substitute:x − y = −3
x − (6− 2x) = −33x − 6 = −3
x = 1
Substituting we obtain y = 4. So, the solution is the point (1, 4).
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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).
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
Applications
Example: Solve the linear system by substitution:
2x + y = 6x − y = −3
2x + y = 6y = 6− 2x now substitute:
x − y = −3
x − (6− 2x) = −33x − 6 = −3
x = 1
Substituting we obtain y = 4. So, the solution is the point (1, 4).
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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).
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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 = −3
x = 1
Substituting we obtain y = 4. So, the solution is the point (1, 4).
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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).
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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).
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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.
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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.
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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.
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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.
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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.
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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.
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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.
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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.
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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)
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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)
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
Applications
Example: Solve the linear system using elimination by addition
3u − 2v = 127u + 2v = 8
3u − 2v = 12+7u + 2v = 8
10u + 0 = 20u = 2
Substituting, we obtain v = −3. So the solution is (2,-3)
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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)
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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)
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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)
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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.
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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.
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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 = 40
4x + 10y = −219x = 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.
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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.
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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.
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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.
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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.
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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.
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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).
The slope of this line is:
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
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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).
The slope of this line is:
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
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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).
The slope of this line is:
m =5.10− 4.80
1.8− 2.0= −0.3
0.2= −3
2
Therefore 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
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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).
The slope of this line is:
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
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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).
The slope of this line is:
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
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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).
The slope of this line is:
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
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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 = 5.85
q = 1.95 billion bushels
Substituting, we find p = −32(1.95) + 7.80 = $4.88
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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)
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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)
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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)
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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)
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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)
Jason Aubrey Math 1300 Finite Mathematics
university-logo
Systems in Two VariablesGraphing
SubstitutionElimination by Addition
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)
Jason Aubrey Math 1300 Finite Mathematics
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