Download Section 6.1 Systems of Two Linear Equations in Two Variables

Section 6.1 Systems of Two Linear Equations in Two Variables
EXAMPLE: Solve the system of equations.
Solution: Begin by solving for in Equation 1.
Next, substitute this expression for y into Equation 2 and solve the resulting single-variable equation for
x.
Finally, you can solve for by back-substituting x = 3 into the equation y = 4 − x, to obtain
The solution is the ordered pair (3, 1). You can check this solution as follows.
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EXAMPLE: Solve the system of equations.
Solution: You can eliminate the y-terms by adding the two equations.
So, x =
3
12
= . By back-substituting into Equation 1, you can solve for y.
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2
The solution is
3 1
, − . You can check the solution algebraically by substituting into the original system.
2 4
2
EXAMPLE: Solve the system of equations.
Solution: You can obtain coefficients of y that differ only in sign by multiplying Equations 1 by 4 and
multiplying Equation 2 by 3.
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From this equation, you can see that x =
= 3. By back-substitution this value of x into Equation 2,
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you can solve for y.
The solution is (3, −2). You can check the solution algebraically by substituting into the original system.
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EXAMPLE: Find all solutions of the system

 2x − y = 5

x + 4y = 7
Solution 1(Substitution Method): We solve for x in the second equation.
x + 4y = 7
x = 7 − 4y
⇐⇒
Now we substitute for x in the first equation and solve for y:
2x − y = 5
2(7 − 4y) − y = 5
14 − 8y − y = 5
14 − 9y = 5
−9y = −9
y=1
Finally, we back-substitute y = 1 into the equation x = 7 − 4y:
x = 7 − 4(1) = 3
Solution 2(Elimination Method): We have


 2x − y = 5
 2x − y = 5
⇐⇒


2x + 8y = 14
x + 4y = 7
⇐⇒
Next we substitute y = 1 into the equation 2x − y = 5:
2x − 1 = 5
⇐⇒
2x = 6
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
 2x − y = 5

9y = 9
⇐⇒
x=3
⇐⇒

 2x − y = 5

y=1
EXAMPLE: Find all solutions of the system

 2x + y = 1

3x + 4y = 14
Solution 1(Substitution Method): We solve for y in the first equation.
2x + y = 1
y = 1 − 2x
⇐⇒
Now we substitute for y in the second equation and solve for x:
3x + 4y = 14
3x + 4(1 − 2x) = 14
3x + 4 − 8x = 14
−5x = 10
x = −2
Finally, we back-substitute x = −2 into the equation y = 1 − 2x:
y = 1 − 2(−2) = 5
Solution 2(Elimination Method): We have


 2x + y = 1
 8x + 4y = 4
⇐⇒


3x + 4y = 14
3x + 4y = 14
⇐⇒

 5x = −10

⇐⇒
3x + 4y = 14

Next we substitute x = −2 into the equation 3x + 4y = 14:
3(−2) + 4y = 14
⇐⇒
−6 + 4y = 14
5
⇐⇒
4y = 20

 x = −2
⇐⇒
3x + 4y = 14
y=5
EXAMPLE: Find all solutions of the system

 2x + 3y = −4

5x − 7y = 1
Solution 1(Substitution Method): We solve for x in the first equation.
2x + 3y = −4
⇐⇒
2x = −4 − 3y
x=−
⇐⇒
4 + 3y
2
Now we substitute for x in the second equation and solve for y:
5x − 7y = 1
−5
4 + 3y
− 7y = 1
2
−5(4 + 3y) − 14y = 2
−20 − 15y − 14y = 2
−29y = 22
y=−
Finally, we back-substitute y = −
22
29
4 + 3y
22
into the equation x = −
:
29
2
4 + 3 − 22
25
29
x=−
=−
2
29
Solution 2(Elimination Method): On the one hand, we have


 10x + 15y = −20
 2x + 3y = −4
=⇒
⇐⇒


10x − 14y = 2
5x − 7y = 1
On the other hand, we have

 2x + 3y = −4

5x − 7y = 1
⇐⇒

 14x + 21y = −28

=⇒
15x − 21y = 3
EXAMPLE: Find all solutions of the system

3
1
7


 −5x + 2y = 2


 1x + 4y = 3
3
5
2
6
29y = −22
⇐⇒
y=−
22
29
29x = −25
⇐⇒
x=−
25
29
EXAMPLE: Find all solutions of the system

1
7
3


 −5x + 2y = 2


 1x + 4y = 3
3
5
2
Solution (Elimination Method): We have


1
7
3


 −5x + 2y = 2
 −6x + 5y = 35
⇐⇒



10x + 24y = 45
 1x + 4y = 3
3
5
2
On the one hand, we have


 −6x + 5y = 35
 −30x + 25y = 175
⇐⇒
=⇒ 97y = 310


10x + 24y = 45
30x + 72y = 135
On the other hand, we have

 −6x + 5y = 35
⇐⇒

10x + 24y = 45

 −144x + 120y = 840

50x + 120y = 225
=⇒
−194x = 615
⇐⇒
⇐⇒
y=
310
97
x=−
615
194
EXAMPLE: Solve the system of linear equations
Solution: Because the coefficients in this system have two decimal places, you can begin by multiplying
each equation by 100 to produce a system with integer coefficients.
Now, to obtain coefficients that differ only by sign, multiply revised Equation 1 by 3 and multiply revised
equation 2 by -2.
So, you can conclude that y =
−322
= 14. Back-substitution this value into revised Equation 2 produces
−23
the following.
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EXAMPLE: Find all solutions of the system

 x+y =1

Answer: No solution.
x+y =2
EXAMPLE: Find all solutions of the system

 3x − 2y = 4

Solution: We have
−6x + 4y = 7

 3x − 2y = 4
⇐⇒


 3x − 2y = 4

 3x − 2y = − 7
2
It follows that the system has no solution (inconsistent).

−6x + 4y = 7
EXAMPLE: Find all solutions of the system

 x+y =1

x+y =1
Answer: The system has infinitely many solutions (dependent).
EXAMPLE: Find all solutions of the system

 8x − 2y = −4
Solution: We have


 8x − 2y = −4

−4x + y = 2
⇐⇒
−4x + y = 2

 8x − 2y = −4

8x − 2y = −4
It follows that the system has infinitely many solutions (dependent).
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