math 1300: section 4-1 review: systems of linear equations in two variables
DESCRIPTION
Lecture over Section 4-1 of Barnett's "Finite Mathematics."TRANSCRIPT
![Page 1: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/1.jpg)
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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
![Page 2: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/2.jpg)
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
![Page 3: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/3.jpg)
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
![Page 4: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/4.jpg)
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
![Page 5: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/5.jpg)
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
![Page 6: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/6.jpg)
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
![Page 7: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/7.jpg)
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
![Page 8: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/8.jpg)
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
![Page 9: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/9.jpg)
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
![Page 10: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/10.jpg)
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
![Page 11: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/11.jpg)
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
![Page 12: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/12.jpg)
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
![Page 13: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/13.jpg)
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
![Page 14: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/14.jpg)
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
![Page 15: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/15.jpg)
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
![Page 16: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/16.jpg)
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
![Page 17: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/17.jpg)
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
![Page 18: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/18.jpg)
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
![Page 19: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/19.jpg)
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
![Page 20: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/20.jpg)
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
![Page 21: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/21.jpg)
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
![Page 22: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/22.jpg)
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
![Page 23: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/23.jpg)
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
![Page 24: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/24.jpg)
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
![Page 25: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/25.jpg)
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
![Page 26: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/26.jpg)
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
![Page 27: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/27.jpg)
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
![Page 28: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/28.jpg)
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
![Page 29: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/29.jpg)
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
![Page 30: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/30.jpg)
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
![Page 31: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/31.jpg)
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
![Page 32: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/32.jpg)
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
![Page 33: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/33.jpg)
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
![Page 34: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/34.jpg)
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
![Page 35: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/35.jpg)
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
![Page 36: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/36.jpg)
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
![Page 37: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/37.jpg)
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
![Page 38: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/38.jpg)
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
![Page 39: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/39.jpg)
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
![Page 40: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/40.jpg)
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
![Page 41: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/41.jpg)
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
![Page 42: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/42.jpg)
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
![Page 43: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/43.jpg)
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
![Page 44: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/44.jpg)
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
![Page 45: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/45.jpg)
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
![Page 46: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/46.jpg)
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
![Page 47: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/47.jpg)
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
![Page 48: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/48.jpg)
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
![Page 49: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/49.jpg)
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
![Page 50: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/50.jpg)
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
![Page 51: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/51.jpg)
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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
![Page 52: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/52.jpg)
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
![Page 53: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/53.jpg)
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
![Page 54: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/54.jpg)
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
![Page 55: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/55.jpg)
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
![Page 56: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/56.jpg)
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
![Page 57: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/57.jpg)
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
![Page 58: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/58.jpg)
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
![Page 59: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/59.jpg)
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
![Page 60: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/60.jpg)
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
![Page 61: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/61.jpg)
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
![Page 62: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/62.jpg)
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
![Page 63: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/63.jpg)
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
![Page 64: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/64.jpg)
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
![Page 65: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/65.jpg)
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
![Page 66: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/66.jpg)
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
![Page 67: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/67.jpg)
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
![Page 68: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/68.jpg)
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
![Page 69: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/69.jpg)
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
![Page 70: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/70.jpg)
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
![Page 71: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/71.jpg)
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
![Page 72: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/72.jpg)
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
![Page 73: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/73.jpg)
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
![Page 74: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/74.jpg)
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
![Page 75: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/75.jpg)
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
![Page 76: Math 1300: Section 4-1 Review: Systems of Linear Equations in Two Variables](https://reader033.vdocuments.net/reader033/viewer/2022051514/5482f0ebb47959f60c8b492c/html5/thumbnails/76.jpg)
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