Download Section 8.1 System of Linear Equations: Substitution and Elimination

Section 8.1 System of Linear Equations: Substitution and Elimination
Identifying Linear Systems
 Linear Systems in two variables
Solution
,
0
0
Consistent/ Independent
Exactly one solution
Two lines cross at one point
1
Consistent/ Dependent
Infinitely many solutions
Two lines are identical
1
Inconsistent
No solution
Two lines are parallel
 Linear Systems in Three Variables
Solution
, ,
0 0
0
0
0 0
Consistent/ Independent
Exactly one solution
Three lines cross at one point
Cheon-Sig Lee
1 0
0 1
Consistent/ Dependent
Infinitely many solutions
Three lines are identical
www.coastalbend.edu/lee
1 0
0 1
Inconsistent
No solution
Three lines are parallel
Page 1
Section 8.1 System of Linear Equations: Substitution and Elimination
Solving Linear Systems by Substitution
 Step 1: Solve either of the equations for one variable in terms of the other variable.
 Step 2: Substitute the expression obtained in step 1 into the other equation, then solve the
resulting equation containing one variable
 Step 3: Back-substitute the value obtained in step 3 into one of the original equations.
Example 1 Solve by the substitution method:
5
4
2
9
3
(Solution 1)
Step 1: Solve either of the equations for one variable in terms of the other variable.
From the second equation,
2
2
3
2
2
3
Step 2: Substitute the expression obtained in step 1 into the other equation, and then solve
resulting equation containing one variable.
5 2
10
5
3
15
6
4
4
4
15
15
6
6
6
9
9
9
9
15
24
24
6
4
←
2
3
Step 3: Back-substitute the value obtained in step 3 into one of the original equations.
or
Using the first equation
Using the second equation
5
5
4
4 4
5
16
16
5
5
5
9
9
9
16
25
25
5
5
←
4
2
2 4
8
8
3
3
3
8
←
4
5
Therefore, the solution is 5, 4 and two lines intersect at 5, 4 .
Cheon-Sig Lee
www.coastalbend.edu/lee
Page 2
Section 8.1 System of Linear Equations: Substitution and Elimination
Solving Linear Systems by Elimination
 Step 1: Rewrite both equations in the form of Ax + By = C.
 Step 2: If necessary, multiply either equation or both equations by appropriate nonzero
numbers so that the x-coefficients or the y-coefficients are same.
 Step 3: Add or Substitute the equations obtained in step 2, then solve the resulting
equation in one variable.
 Step 4: Back-substitute the value obtained in step4 into one of the original equations.
Example 2: Solve by the elimination method:
5
4
2
9
3
(Solution 2)
Step 1: Rewrite both equations in the form of Ax + By = C. (Skip)
Step 2: If necessary, multiply either equation or both equations by appropriate nonzero
numbers so that the x-coefficients or the y-coefficients are same.
Multiply the second equation by 5. So, we have
5
5
4
10
9
15
←5∙
5∙2
3∙5
Step 3: Add or Substitute the equations obtained in step 2, then solve the resulting equation in
one variable
5
5
4
10
6
6
6
9
15
24
24
6
4
Step 4: Back-substitute the value obtained in step 3 into one of the original equations
or
Using the first equation
Using the second equation
5
5
4
4 4
5
16
16
5
5
5
9
9
9
16
25
25
5
5
←
4
2
2 4
8
8
3
3
3
8
←
4
5
Therefore, the solution is 5, 4 and two lines intersect at 5, 4 .
Cheon-Sig Lee
www.coastalbend.edu/lee
Page 3
Section 8.1 System of Linear Equations: Substitution and Elimination
Solving Linear Systems by Graphing (TI-83)
 Step1: Rewrite both equations in the form of y = mx + b.
5
4
2
9⇒
3⇒
9
4
3
2
5
4
1
2
 Step2: Type the resulted equations in your calculator.
 Step3: Graph and then find intercept.
 TI-83/84: graph and then 2nd TRACE .
 TI-89: graph and then F5
Solving Linear Systems by Matrix Method (TI-83)
 Step1: Rewrite both equations in the augmented matrix form
5
4
9
5
4 9
⟹
2
3
1
2
3
 Step2: Type the matrix obtained in step1 in your calculator
 TI-83/84: 2nd x–1 > Edit > Dimension 2×3 >
5
1
4
2
9
> 2nd MODE
3
> 2nd x–1 > MATH > rref > 2nd x–1 > [A]
 TI-89: 2nd 5 > Matrix > rref ([5,4,9;1,-2,-3])
TI-83: Matrix Method
2nd x–1 > Edit > Dimension 2×3 >
Cheon-Sig Lee
5
1
4
2
9
–1
> 2nd MODE > 2nd x–1 > MATH > rref > 2nd x > [A]
3
www.coastalbend.edu/lee
Page 4
Section 8.1 System of Linear Equations: Substitution and Elimination
Exercises
1.
(Solution 1)
Elimination Method
Step 1:
① ③⇒
3
2
2
3
2
14
Step 4: Substitute 2 for x in the equation ④or⑤, then
solve for the variable z. From the equation ④,
6
8
2
8
←
2
∙∙∙∙∙∙∙ ④
2 2
8
14
4
8
4
4
4
10 ∙∙∙∙∙∙∙ ⑤
Step 5: Substitute 2 for x and 4 for z into one of the
original equations. From the equation ②,
10
40
2
2 ←
2,
4
30
2 2
4
2
30
Therefore, the solution is 2, 2, 4
15
2
Step 2: ①
2∗②⇒
4
Step 3: ⑤
5∗④⇒
Matrix Method

Augmented Matrix:
2
2
5
3
2
5
5
10
5
5
15
15
15
1
2
2 1
3 2
3 14
1
2
3
6
x
y
z
⇒
Augmented Matrix
Cheon-Sig Lee
www.coastalbend.edu/lee
Page 5
Section 8.1 System of Linear Equations: Substitution and Elimination
2.
(Solution 2)
Elimination Method
Step 1:
① ②⇒
4
3
5
∙∙∙∙∙∙∙ ④
0
45
0
45 ∙∙∙∙∙∙∙ ⑤
The equation ⑤ has no solution. Thus the system is inconsistent.
Step 2: ③
5∗④⇒
Matrix Method

Augmented Matrix:
25
25
2
4
5
9
10
10
1
1
1 4
4 3 1 5
3 2 0 0
⇒
No Solution
Augmented Matrix
Cheon-Sig Lee
www.coastalbend.edu/lee
Page 6
Section 8.1 System of Linear Equations: Substitution and Elimination
3.
(Solution 3)
Elimination Method
Step 1:
① ②⇒
2
2
Step 2: ② ∗ 5
③⇒
5
5
3
4
10
2
8
7
5
15
17
32
∙∙∙∙∙∙∙ ④
35
5
40
Step 4: Write the solution in terms of one variable.
Write x and y in terms of z because all
equations from ①to⑤havethevariablez .
From the equation ④,
4
5
4
4 ∙∙∙∙∙∙∙ ⑤
32
40
32
40
0 0
∙∙∙∙∙∙∙ ⑥
The equation ⑥ has infinitely many solutions. Thus, the
system has consistent and dependent.
Step 3: ④ ∗ 8
⑤⇒
8
8
Therefore, there are infinitely many solutions and the solution set is
From the equation ①,
2
4
, ,
|
5
3,
5
4
2 ←
2
4
5for ∀ z
Matrix Method
5
3⇒
4
5⇒
Infinitely many solutions
⇒
Augmented Matrix
Cheon-Sig Lee
www.coastalbend.edu/lee
Page 7
5
Section 8.1 System of Linear Equations: Substitution and Elimination
4.
(Solution 4)
Step 1:
①
Step 2: ① ∗ 2
Step 3:
④
②⇒
③⇒
3
6
6
3
⑤⇒
3
5
3
4
4
2
6
3
3
∙∙∙∙∙∙∙ ④
6
4
3
8
8
2
12
14
2
∙∙∙∙∙∙∙ ⑤
3
3
2
2
3
2
0 1
∙∙∙∙∙∙∙ ⑥
The equation ⑥ has no solution. Thus the system is inconsistent.
Matrix Method
⇒
No Solution
Augmented Matrix
Cheon-Sig Lee
www.coastalbend.edu/lee
Page 8
Section 8.1 System of Linear Equations: Substitution and Elimination
5.
(Solution 5)
Step 1:
①
6
②⇒
4
Step 2: ① ∗ 4
③⇒
Step 3: ④ ∗ 3
⑤⇒
2
5
4
4
2
4
7
3
4
4
3
15
3
3
3
12
12
12
24
30
6
Step 4: Substitute 1 for x in the equation ④or⑤, then
solve for the variable y. From the equation ④,
5
2 ←
1
5 1
2
∙∙∙∙∙∙ ④
5
2
5
5
3
3
∙∙∙∙∙∙ ⑤
1
1
6
6
12
12
12
Step 5: Substitute 1 for x and 3 for z into one of the
original equations. From the equation ①,
6 ←
1,
3
1 3
6
4
6
4
4
2 ⇒
Matrix Method
⇒
x
y
z
Augmented Matrix
Cheon-Sig Lee
www.coastalbend.edu/lee
Page 9
Section 8.1 System of Linear Equations: Substitution and Elimination
6.
(Solution 6)
Step 1:
①
2
4
②⇒
3
Step 2: ② ∗ 2
③⇒
Step 3: ④ ∗ 2
⑤⇒
6
3
4
2
8
4
3
2
2
4
8
3
4
4
5
5
5
Matrix Method
⇒
3
6
9
Step 4: Substitute 2 for x in the equation ④or⑤, then
solve for the variable y. From the equation ④,
4
2
9
←
2
∙∙∙∙∙∙ ④
4 2
2
9
12
8 2
9
4
8
8
8 ∙∙∙∙∙∙ ⑤
2
1
2
1
⟹
18
2
2
8
Step 5: Substitute 1 for x and 3 for z into one of the
original equations. From the equation ①,
10
1
10
2
3 ←
2,
2
1
5
3
2 2
2
2 1
3
1
3
1
1
2
2
⟹
1
1
x
y
z
Augmented Matrix
Cheon-Sig Lee
www.coastalbend.edu/lee
Page 10