5 – 3:
Solving a System of Two Equations by Elimination
Selected Worked Homework Problems
Solve each system by the elimination method. List your answers as an ordered pair.
1.
Equation A ⎧ 3x + 5y = 2
⎨
Equation B ⎩−3x + y = −14
!
If you add the left sides of Equation A and Equation B together
and add the right sides of Equation A and Equation B together
the x terms add to zero and you will have eliminated the x terms.
You now have a new equation with only the y variable.
Add Equation A and Equation B to eliminate the x terms
3x + 5y = 2
−3x + 1y = −14
6y = −12
Solve for y
y = −2
Plug y = −2 into either equation A or B
and solve for x
Equation A
3x + 5(−2) = 2
3x − 10 = 2
3x = 12
x=4
Answer: (4,−2)
check:
Equation A
3x + 5y = 2
3(4) + 5(−2) = 2
12 − 10 = 2
Math 100 !
Equation B
− 3x + 1y = −14
− 3(4) + 1(−2) = −14
− 12 − 2 = −14
Section 5 – 3 HW WKD
!
© 2016 Eitel
2.
Equation A ⎧ 3x − 2y = −11
⎨
Equation B ⎩−5x + 2y = 21
If you add the left sides of Equation A and Equation B together
and add the right sides of Equation A and Equation B together
the y terms add to zero and you will have eliminated the y terms.
You now have a new equation with only the x variable.
Add Equation A and Equation B to eliminate the y terms
3x − 2y = −11
−5x + 2y = 21
−2x = 10
Solve for y
x = −5
Plug x = −5 into either equation A or B
and solve for y
Equation A
3x − 2y = −11
3(−5) − 2y = −11
−15 − 2y = −11
−2y = 4
y = −2
Answer: (−5,−2)
check:
Equation A
3x − 2y = −11
3(−5) − 2(−2) = −11
−15 + 4 = −11
Math 100 !
Equation B
− 5x + 2y = 21
− 5(−5) + 2(−2) = 21
25 − 4 = 21
Section 5 – 3 HW WKD
!
© 2016 Eitel
3.
Equation A ⎧ 2x − y = −3
⎨
Equation B ⎩−2x + 3y = −9
If you add the left sides of Equation A and Equation B together
and add the right sides of Equation A and Equation B together
the x terms add to zero and you will have eliminated the x terms.
You now have a new equation with only the y variable.
Add Equation A and Equation B to eliminate the x terms
2x − y = −3
−2x + 3y = −9
2y = −12
Solve for y
y = −6
Plug y = −6 into either equation A or B
and solve for x
Equation A
2x − (y) = −3
2x − (−6) = −3
2x + 6 = −3
2x = −9
−9
x=
2
⎛ -9 ⎞
Answer:
,−6
⎝2
⎠
check:
Equation A
2x − y = −3
Equation B
− 2x + 3y = −9
⎛ −9 ⎞
2
− (−6) = −3
⎝ 2⎠
−2
−9 + 6 = −3
9 − 18 = −9
Math 100 !
⎛ −9 ⎞
+ 3(−6) = −9
⎝ 2⎠
Section 5 – 3 HW WKD
!
© 2016 Eitel
4.
Equation A ⎧ 4 x − 5y = −3
⎨
Equation B ⎩−4 x + 5y = 1
If you add the left sides of Equation A and Equation B together
and add the right sides of Equation A and Equation B together
the y terms add to zero and you will have eliminated the y terms.
You now have a new equation with only the x variable.
Add Equation A and Equation B to eliminate the y terms
4x − 5y = −3
−4x + 5y = 1
0 = −2
STOP: Both the x and y terms canceled out
and the remaining equation 0 = −2 is false
The lines are parallel,they have no common points.
Answer: No Solution
Math 100 !
Section 5 – 3 HW WKD
!
© 2016 Eitel
5.
Equation A ⎧ 2x − y = −4
⎨
Equation B ⎩−2x + y = 4
!
If you add the left sides of Equation A and Equation B together
and add the right sides of Equation A and Equation B together
the y terms add to zero and you will have eliminated the y terms.
You now have a new equation with only the x variable.
Add Equation A and Equation B to eliminate the y terms
2x − y = −4
−2x + y = 4
0 = −2
STOP: Both the x and y terms canceled out
and the remaining equation 0 = 0 is true
Both equations describe the same line
any point on 2x − y = −4
would also be on − 2x + y = 4
Answer: All Points on 2x − y = −4
or
Answer: All Points on − 2x + y = 4
either one of the above is correct
Math 100 !
Section 5 – 3 HW WKD
!
© 2016 Eitel
6.
Equation A ⎧ − x − 3y = −6
⎨
Equation B ⎩−2x + 3y = 0
If you add the left sides of Equation A and Equation B together
and add the right sides of Equation A and Equation B together
the y terms add to zero and you will have eliminated the y terms.
You now have a new equation with only the x variable.
Add Equation A and Equation B to eliminate the y terms
− x − 3y = −6
−2x + 3y = 0
−3x = −6
Solve for y
x=2
Plug x = 2 into either equation A or B
and solve for y
Equation A
− x − 3y = −6
−(2) − 3y = −6
−2 − 3y = −6
−3y = −4
4
y=
3
⎛ 4⎞
Answer: 2,
⎝ 3⎠
check:
Equation A
− x − 3y = −6
⎛ 4⎞
−(2) − 3
= −6
⎝ 3⎠
−2 − 4 = −6
Math 100 !
Equation B
− 2x + 3y = 0
⎛ 4⎞
− 2(2) + 3
)=0
⎝ 3⎠
−4+4=0
Section 5 – 3 HW WKD
!
© 2016 Eitel
7.
Equation A ⎧2x + 5y = 9
!
⎨
Equation B ⎩3x + y = 7
8.
Equation A ⎧ 2x − 3y = 5
⎨
Equation B ⎩−4x + 2y = 2
Multiply Equation B by − 5 so that
Equation A has 5x and
Equation B has − 5x
Multiply Equation A by 2 so that
Equation A has 4x and
Equation B has − 4 x
⎧2x + 5y = 9
⎨
−5 ⎩3x + y = 7
2 ⎧2x − 3y = 5
⎨
⎩−4 x + 2y = 2
2x + 5y = 9
−15x − 5y = −35
4 x − 6y = 10
−4 x + 2y = 2
− 13x = −26
x =2
− 4y = 12
y = −3
!
!
Plug x = 2 into either equation A or B
and solve for y
Plug y = −3 into
either equation A or B
and solve for x
Equation A
2(2) + 5y = 9
4 + 5y = 9
5y = 5
y =1
Answer: (2,1 )
Math 100 !
Equation A
2(x) − 3(−3) = 5
2x + 9 = 5
2x = −4
x = −2
Answer: (− 2,−3 )
Section 5 – 3 HW WKD
!
© 2016 Eitel
9.
Equation A ⎧3x + y = −7
!
⎨
Equation B ⎩ x + 2y = 6
10.
⎧−4 x + 5y = −16
⎨
⎩8x + y = −1
Multiply Equation B by − 3 so that
Equation A has 3x and
Equation B has − 3x
Multiply Equation A by 2 so that
Equation A has − 8x and
Equation B has 8x
⎧3x + y = −7
⎨
−3 ⎩ x + 2y = 6
2 ⎧−4 x + 5y = −16
⎨
⎩8x + y = −1
3x + y = −7
−3x − 6y = −18
−8x + 10y = −32
8x + y = −1
− 5y = −25
y =5
11y = −33
y = −3
!
Plug y = 5 into
either equation A or B
and solve for x
Plug y = −3 into either
equation A or B
and solve for x.
Equation A
Equation A
−4(x) + 5(−3) = −16
−4 x − 15 = −16
−4 x = −1
1
x=
4
⎛1 ⎞
Answer:
,−3
⎝4 ⎠
3(x) + ( y ) = −7
3(x) + (5) = −7
3x + 5 = −7
3x = −12
x = −4
Answer: (− 4,5 )
NOTE: You could have multiplied!
Equation A by –2 instead of!
multiplying Equation B by –3!
This would eliminate the y terms !
and you would then solve for x.!
NOTE: You could have multiplied
Equation B by –2 instead of
multiplying Equation B by –5
This would eliminate the y terms
and you would then solve for x.
You get the same answer in either case.
Math 100 !
Section 5 – 3 HW WKD
!
© 2016 Eitel
14.
Equation A ⎧2x + 3y = −8
⎨
Equation B ⎩5x + 4 y = −34
You must mutiply both rows by
different numbers to eliminate a variable
Multiply Equation A by 5
Multiply Equation B by − 2
to eliminate the x terms
5 ⎧2x + 3y = −8
⎨
−2 ⎩5x + 4 y = −34
10x + 15y = −40
−10x − 8y = 68
7y = 28
y=4
Now add the two equations
Solve for y
Plug y = 4 into either equation A or B
and solve for x
Equation A
2x + 3(y) = −8
2x + 3(4) = −8
2x + 12 = −8
2x = −20
x = −10
Answer: (−10,4 )
NOTE: You could have multiplied Equation A by –2!
instead of multiplying Equation B by –3!
This would eliminate the y terms !
and you would then solve for x.!
!
You get the same answer in either case.
Math 100 !
Section 5 – 3 HW WKD
!
© 2016 Eitel
15.
Equation A ⎧3x − 5y = 11
!
⎨
Equation B ⎩4 x + 3y = 5
You must mutiply both rows by
different numbers to eliminate a variable
Multiply Equation A by 3
Multiply Equation B by 5
to eliminate the y terms
3 ⎧ 3x − 5y = 11
⎨
5 ⎩ 4x + 3y = 5
9x − 15y = 33
20x + 15y = 25
29x = 58
x=2
Now add the two equations
Solve for y
Plug x = 2 into either equation A or B
and solve for y
Equation A
3(x) − 5(y) = 11
3(2) − 5(y) = 11
6 − 5y = 11
−5y = 5
y = −1
Answer: ( 2,−1)
Math 100 !
Section 5 – 3 HW WKD
!
© 2016 Eitel
21.
x 3y −1
Equation A ⎧ −
⎪3 4 = 2
⎨
x y 3
Equation B ⎪ + =
⎩6 8 4
!
Multiply Equation A by 12
Multiply Equation B by 24
to eliminate the fractions
x 3y −1
12 ⎧ −
⎪3 4 = 2
⎨
x y 3
24 ⎪ + =
⎩6 8 4
to get 2 new equations
Equation C ⎧4 x − 9y = −6
⎨
Equation D ⎩4 x + 3y = 18
which you now can solve
Multiply Equation A by − 1
to eliminate the x terms
−1⎧4 x − 9y = −6
⎨
⎩4 x + 3y = 18
−4 x + 9y = 6
4 x + 3y = 18
12y = 24
y=2
Now add the two equations
Solve for y
Plug y = 2 into either equation A or B
and solve for x
Equation A
4(x) − 9(y) = −6
4(x) − 9(2) = −6
4 x − 18 = −6
4 x = 12
x=3
Answer: ( 3,2)
Math 100 !
Section 5 – 3 HW WKD
!
© 2016 Eitel
22.
Equation A ⎧ x − y = 3
⎪⎪ 2 4 4
⎨
x y 7
Equation B ⎪⎪ + =
⎩2 6 6
Multiply Equation A by 4
Multiply Equation B by 6
to eliminate the fractions
x y 3
4⎧ − =
⎪2 4 4
⎨
x y 7
6⎪ + =
⎩2 6 6
to get 2 new equations
Equation C ⎧2x − y = 3
⎨
Equation D ⎩3x + y = 7
which you now can solve
Add Equation C and Equation D to eliminate
the y terms and solve for x
Equation C ⎧2x − y = 3
⎨
Equation D ⎩ 3x + y = 7
5x = 10
x=2
Plug x = 2 into either equation C or D
and solve for x
Equation C
2(x) − (y) = 3
2(2) − (y) = 3
4−y= 3
−y = −1
y =1
Answer: ( 2,1)
Math 100 !
Section 5 – 3 HW WKD
!
© 2016 Eitel