Download HW Section 5

5 – 2:
Solving a System of Two Equations by Substitution
Selected Worked Homework Problems
Solve each system by the substitution method.
1.
Equation A ⎧3x + 5y = 26
⎨
Equation B ⎩ y = 2x
!
Equation B has been solved for y.
y = 2x
Substitute the expression (2x)
into Equation A in place of y.
Equation A with 2x subtituted for y
3x + 5(2x) = 26
Solve for x
3x + 5(2x) = 26
3x + 10x = 26
13x = 26
x =2
plug x = 4 into Equation B to find y
y = 2(2) = 4
Answer: (2,4)
check:
Equation A
3(x) + 5(y) = 26
3(2) + 5(4) = 26
6 + 20 = 26
26 = 26
Math 100 !
Equation B
y = 2(x)
4 = 2(2)
4=4
Section 5 – 2 HW WKD!
© 2016 Eitel
2.
Equation A ⎧2x + 3y = 12
⎨
Equation B ⎩ y = x − 1
Equation B has been solved for y.
y = x −1
Substitute the expression (x − 1)
into Equation A in place of y.
Equation A with 2x subtituted for y
2x + 3(x − 1) = 12
Solve for x
2x + 3(x − 1) = 12
2x + 3x − 3 = 12
5x = 15
x=3
plug x = 3 into Equation B to find y
y = (3) − 1 = 2
Answer: (3,2)
check:
Equation A
2(x) + 3(y) = 12
2(3) + 3(2) = 12
6 + 6 = 12
12 = 12
Math 100 !
Equation B
y = (x) − 1
2 = 3−1
2=2
Section 5 – 2 HW WKD!
© 2016 Eitel
3.
Equation A ⎧−3x + 2y = 16
⎨
Equation B ⎩ y = −2x + 1
!
Equation B has been solved for y.
y = −2x + 1
Substitute the expression (−2x + 1 )
into Equation A in place of y.
Equation A with 2x subtituted for y
−3x + 2(−2x + 1) = 16
Solve for x
−3x + 2(−2x + 1) = 16
−3x − 4 x + 2 = 16
−7x = 14
x = −2
plug x = −2 into Equation B to find y
y = −2(−2) + 1 = 5
Answer: (−2,5)
check:
Equation A
−3(x) + 2(y) = 16
−3(−2) + 2(5) = 16
6 + 10 = 16
16 = 16
Math 100 !
Equation B
y = −2(x) + 1
− 3 = −2(−2) + 1
5 = 4 +1
5=5
Section 5 – 2 HW WKD!
© 2016 Eitel
4.
Equation A ⎧−3x − y = −9
⎨
Equation B ⎩ x = − y + 1
Equation B has been solved for x.
Substitute the expression x is equal to (− y + 1)
into Equation A in place of the x.
Equation A with − y + 1 subtituted for x
−3(− y + 1) − y = −9
Solve for y
−3(− y + 1) − y = −9
3y − 3 − y = −9
2y − 3 = −9
2y = −6
y = −3
plug y = 6 into Equation B to find x
x = −(−3) + 1 = 3 + 1 = 4
Answer: (4,−3)
check:
Equation A
−3(x) − (y) = −9
−3(4) − (−3) = −9
−12 + 3 = −9
−9 = −9
Math 100 !
Equation B
4 = −(y) + 1
4 = −(−3) + 1
4 = 3 +1
4=4
Section 5 – 2 HW WKD!
© 2016 Eitel
5.
Equation A ⎧−2x − 6y = −18
⎨
Equation B ⎩ x = −2y + 3
!
Equation B has been solved for x.
Substitute the expression x is equal to (−2y + 3)
into Equation A in place of the x.
Equation A with − 2y + 3 subtituted for x
−2(−2y + 3) − 6y = −18
Solve for y
−2(−2y + 3) − 6y = −18
4y − 6 − 6y = −18
−2y − 6 = −18
−2y = −12
y=6
plug y = 6 into Equation B to find x
x = −2(6) + 3
x = 12 + 3
x = −9
Answer: (−9,6)
check:
Equation A
−2(x) − 6(y) = −18
−2(−9) − 6(6) = −18
18 − 36 = −18
−18 = −18
Math 100 !
Equation B
− 9 = −2(y) + 3
− 9 = −2(6) + 3
− 9 = −12 + 3
− 9 = −9
Section 5 – 2 HW WKD!
© 2016 Eitel
6.
Equation A ⎧6x + 3y = 16
⎨
Equation B ⎩ y = −2x + 1
Equation B has been solved for y.
y = −2x + 1
Substitute the expression (−2x + 1 )
into Equation A in place of y.
Equation A with 2x subtituted for y
6x + 3(−2x + 1) = 16
Solve for x
6x + 3(−2x + 1) = 16
6x − 6x + 3 = 16
−3 = 16
Stop: The x term canceled out and
the remaining eqution − 3 = −6 is false
The lines are parallel,
they have no common points
Answer: No Solution
Math 100 !
Section 5 – 2 HW WKD!
© 2016 Eitel
8.
Equation A ⎧−2x − 4y = −6
⎨
Equation B ⎩ x = −2y + 3
Equation B has been solved for x.
Substitute the expression x is equal to (−2y + 3)
into Equation A in place of y.
Equation A with 3y − 4 subtituted for x
−2(−2y + 3) − 4y = −6
Solve for y
−2(−2y + 3) − 4y = −6
4y − 6 − 4y = −6
−6 = −6
Stop: The x term canceled out and
the remaining eqution − 6 = −6 is true
Both equations describe the same line
any point on − 2x − 4y = −6
would also be on x = −2y + 3
Answer: All Points on − 2x − 4y = −6
or
Answer: All Points on x = −2y + 3
either one of the above is correct
Math 100 !
Section 5 – 2 HW WKD!
© 2016 Eitel
15.
Equation A ⎧2x − 3y = 8
⎨
Equation B ⎩ x = 6y − 2
Equation B has been solved for x.
Substitute the expression x is equal to (6y − 2)
into Equation A in place of y.
Equation A with 6y − 2 subtituted for x
2(6y − 2) − 3y = 8
Solve for y
2(6y − 2) − 3y = 8
12y − 4 − 3y = 8
9y = 12
12 4
y=
=
9 3
plug y =
4
into Equation B to find x
3
⎛ 4⎞
x=6
−2= 8−2= 6
⎝ 3⎠
Answer:
⎛ 4⎞
6,
⎝ 3⎠
check:
Equation A
2(x) − 3(y) = 8
Equation B
x = 6(y) − 2
⎛ 4⎞
2(6) − 3
=8
⎝ 3⎠
⎛ 4⎞
6=6
−2
⎝ 3⎠
12 − 4 = 8
8=8
6= 8−2
6=6
Math 100 !
Section 5 – 2 HW WKD!
© 2016 Eitel
16.
Equation A ⎧2x + 9y = 4
⎨
Equation B ⎩ x = −3y + 1
Equation B has been solved for x.
Substitute the expression x is equal to (−3y + 1)
into Equation A in place of y.
Equation A with 6y − 2 subtituted for x
2(−3y + 1) + 9y = 4
Solve for y
2(−3y + 1) + 9y = 4
−6y + 2 + 9y = 4
3y = 2
2
y=
3
2
into Equation B to find x
3
⎛ 2⎞
x = −3
= 1 = −2 + 1 = −1
⎝ 3⎠
plug y =
Answer:
⎛ 2⎞
−1,
⎝ 3⎠
check:
Equation A
2(x) + 9(y) = 4
Equation B
x = −3(y) + 1
⎛ 2⎞
2(−1) + 9
=4
⎝ 3⎠
⎛ 2⎞
− 1 = −3
+1
⎝ 3⎠
−2 + 6 = 4
4=4
Math 100 !
− 1 = −2 + 1
− 1 = −1
Section 5 – 2 HW WKD!
© 2016 Eitel