Solving System of Two equations
Graphical Method
y =x+1
(1)
y = −x + 5
(2)
Solving System of Two equations
Graphical Method
y =x+1
(1)
y = −x + 5
(2)
are the same equations as:
x − y = −1
(3)
x+y =5
(4)
Solving System of Two equations
Graphical Method
Graphical Method
Solve by graphing equations and finding all common points.
+6.8
y
+6.6
+6.4
+6.2
+6.0
+5.8
+5.6
+5.4
+5.2
+5.0
+4.8
+4.6
+4.4
+4.2
+4.0
+3.8
+3.6
+3.4
+3.2
+3.0
+2.8
+2.6
+2.4
+2.2
+2.0
+1.8
+1.6
+1.4
+1.2
+1.0
+0.8
+0.6
+0.4
+0.2
−1.6
−1.2
−0.8
−0.4
−0.2
−0.4
−0.6
−0.8
+0.4
+0.8
+1.2
+1.6
+2.0
+2.4
+2.8
+3.2
+3.6
x
Solving System of Two equations
Substitution Method (1)
SubstitutionMethod
x − y = −1
(5)
x+y =5
(6)
Substitution Method
solve for x in first equation
Solving System of Two equations
Substitution Method (1)
SubstitutionMethod
x − y = −1
(5)
x+y =5
(6)
Substitution Method
solve for x in first equation
use to eliminate x in second equation
Solving System of Two equations
Substitution Method (1)
SubstitutionMethod
x − y = −1
(5)
x+y =5
(6)
Substitution Method
solve for x in first equation
use to eliminate x in second equation
solve for remaining variable
Solving System of Two equations
Substitution Method (1)
SubstitutionMethod
x − y = −1
(5)
x+y =5
(6)
Substitution Method
solve for x in first equation
use to eliminate x in second equation
solve for remaining variable
use found variable in any equation to find other variable.
Solving System of Two equations
Substitution Method (2)
Substitution Method:
x − y = −1
(7)
x+y =5
(8)
Solving System of Two equations
Substitution Method (2)
x − y = −1
(7)
x+y =5
(8)
Substitution Method: From first equation
x=y−1
(9)
Solving System of Two equations
Substitution Method (2)
x − y = −1
(7)
x+y =5
(8)
Substitution Method: From first equation
x=y−1
(9)
substitute for x in second equation gives:
(y − 1) + y = 5
(10)
2y = 6
(11)
y=3
(12)
Solving System of Two equations
Substitution Method (2)
x − y = −1
(7)
x+y =5
(8)
Substitution Method: From first equation
x=y−1
(9)
substitute for x in second equation gives:
(y − 1) + y = 5
(10)
2y = 6
(11)
y=3
(12)
Now back substitute to find x:
x=y−1=3−1=2
Giving (x, y) = (2, 3) as the only solution.
(13)
Solving System of Two equations
Elimination Method
Elimination Method
x − y = −1
(14)
x + y = 5 Add to eliminate y
(15)
2x + 0 = 4
(16)
x=2
(17)
Solving System of Two equations
Elimination Method
Elimination Method
x − y = −1
(14)
x + y = 5 Add to eliminate y
(15)
2x + 0 = 4
(16)
x=2
(17)
Back Substitute to get
(18)
2 − y = −1
(19)
y=3
(20)
⇒ (x, y) = (2, 3)
(21)
Gauss-Jordan Elimination
new example
Gauss − Jordan Elimination
2x + 3y = 8
(22)
6x − 2y = 2
(23)
Gauss-Jordan Elimination
new example
Gauss − Jordan Elimination
2x + 3y = 8
(22)
6x − 2y = 2
(23)
Use the first elementary row operation: interchange two rows
2x + 3y = 8
6x − 2y = 2
R1 ↔ R2
(24)
(25)
Gauss-Jordan Elimination
new example
Gauss − Jordan Elimination
2x + 3y = 8
(22)
6x − 2y = 2
(23)
Use the first elementary row operation: interchange two rows
R1 ↔ R2
2x + 3y = 8
6x − 2y = 2
(24)
(25)
Gives Equivalent System:
6x − 2y = 2
(26)
2x + 3y = 8
(27)
Gauss-Jordan Elimination
6x − 2y = 2
(28)
2x + 3y = 8
(29)
Gauss-Jordan Elimination
6x − 2y = 2
(28)
2x + 3y = 8
(29)
Use the second elementary row operation: Multiply a row by a
number
6x − 2y = 2
2x + 3y = 8
R1 →
1
2
R1
(30)
(31)
Gauss-Jordan Elimination
6x − 2y = 2
(28)
2x + 3y = 8
(29)
Use the second elementary row operation: Multiply a row by a
number
6x − 2y = 2
R1 →
2x + 3y = 8
1
2
R1
(30)
(31)
Gives Equivalent System:
3x − y = 1
(32)
2x + 3y = 8
(33)
Gauss-Jordan Elimination
3x − y = 1
(34)
2x + 3y = 8
(35)
Gauss-Jordan Elimination
3x − y = 1
(34)
2x + 3y = 8
(35)
Use the third elementary row operation: add to a row some
multiple of another row
3x − y = 1
2x + 3y = 8
R1 → R1 + () · R2
(36)
(37)
Gauss-Jordan Elimination
3x − y = 1
(34)
2x + 3y = 8
(35)
Use the third elementary row operation: add to a row some
multiple of another row
3x − y = 1
R1 → R1 + () · R2
2x + 3y = 8
3x − y = 1
2x + 3y = 8
(36)
(37)
R1 → R1 + (−1) · R2
(38)
(39)
Gauss-Jordan Elimination
3x − y = 1
(34)
2x + 3y = 8
(35)
Use the third elementary row operation: add to a row some
multiple of another row
3x − y = 1
R1 → R1 + () · R2
2x + 3y = 8
3x − y = 1
(36)
(37)
R1 → R1 + (−1) · R2
2x + 3y = 8
(38)
(39)
Gives Equivalent System:
x − 4y = −7
2x + 3y = 8
(40)
(41)
Gauss-Jordan Elimination
x − 4y = −7
2x + 3y = 8
(42)
(43)
Gauss-Jordan Elimination
x − 4y = −7
2x + 3y = 8
(42)
(43)
Use the third elementary row operation
x − 4y = −7
2x + 3y = 8
(44)
R2 → R2 + (−2) · R1
(45)
Gauss-Jordan Elimination
x − 4y = −7
2x + 3y = 8
(42)
(43)
Use the third elementary row operation
x − 4y = −7
2x + 3y = 8
(44)
R2 → R2 + (−2) · R1
(45)
Gives Equivalent System:
x − 4y = −7
0x + 11y = 22
(46)
(47)
Gauss-Jordan Elimination
x − 4y = −7
0x + 11y = 22
System is in Triangular Form or Row-Echelon Form
(48)
(49)
Gauss-Jordan Elimination
x − 4y = −7
0x + 11y = 22
System is in Triangular Form or Row-Echelon Form
At this point typically use one of two methods to continue:
1
Gaussian Elimination
(48)
(49)
Gauss-Jordan Elimination
x − 4y = −7
0x + 11y = 22
System is in Triangular Form or Row-Echelon Form
At this point typically use one of two methods to continue:
1
Gaussian Elimination
use last row to solve for y
(48)
(49)
Gauss-Jordan Elimination
x − 4y = −7
0x + 11y = 22
System is in Triangular Form or Row-Echelon Form
At this point typically use one of two methods to continue:
1
Gaussian Elimination
use last row to solve for y
back-substitute to solve for x
(48)
(49)
Gauss-Jordan Elimination
x − 4y = −7
0x + 11y = 22
System is in Triangular Form or Row-Echelon Form
At this point typically use one of two methods to continue:
1
Gaussian Elimination
use last row to solve for y
back-substitute to solve for x
2
Gauss-Jordan Elimination
(48)
(49)
Gauss-Jordan Elimination
x − 4y = −7
0x + 11y = 22
System is in Triangular Form or Row-Echelon Form
At this point typically use one of two methods to continue:
1
Gaussian Elimination
use last row to solve for y
back-substitute to solve for x
2
Gauss-Jordan Elimination
continue to reduce to Reduced Row Echelon Form (rref)
(48)
(49)
Gauss-Jordan Elimination
x − 4y = −7
0x + 11y = 22
System is in Triangular Form or Row-Echelon Form
At this point typically use one of two methods to continue:
1
Gaussian Elimination
use last row to solve for y
back-substitute to solve for x
2
Gauss-Jordan Elimination
continue to reduce to Reduced Row Echelon Form (rref)
solve for leading variables in terms of non-leading variables.
(48)
(49)
Gauss-Jordan Elimination
Continue with Gauss-Jordan Elimination
x − 4y = −7
0x + 11y = 22
x − 4y = −7
(50)
R2 →
1
R2
(51)
R1 → R1 + (4) · R2
(52)
11
0x + y = 2
(53)
x − 0y = 1
(54)
0x + y = 2
(55)
Gauss-Jordan Elimination
Continue with Gauss-Jordan Elimination
x − 4y = −7
0x + 11y = 22
x − 4y = −7
(50)
R2 →
1
R2
(51)
R1 → R1 + (4) · R2
(52)
11
0x + y = 2
(53)
x − 0y = 1
(54)
0x + y = 2
(55)
Giving (x, y) = (1, 2)