See the attached file for questioins
Thehonest
s1.docxI) Is the ordered pair (1, -1) a solution of the given linear system?
2x – 3y = 5
6x + y = 1
Substituting x = 1 and y = -1 into the first equation gives:
2(1) – 3(-1) = 5
2 + 3 = 5
5 = 5 Checks.
Substituting x = 1 and y = -1 into the second equation gives:
6(1) + (-1) = 1
6 – 1 = 1
5 = 1 Does NOT check.
Since (1, -1) does not satisfy the second equation of the system, it is not a solution.
II) Solve by the Substitution Method
3x + y = 7
4x + 3y = 1
Solving the first equation for y gives:
y = 7 – 3x
Substituting the right hand side of this equation in place of y in the second equation of the system gives:
4x + 3(7 – 3x) = 1
4x + 21 – 9x = 1
9x – 4x = 21 – 1
5x = 20
x = 20/5
x = 4
Substituting this value of x in the equation y = 7 – 3x gives:
y = 7 – 3(4) = 7 – 12 = -5
Solution: x = 4, y = -5
III) Find the solution of the pair of equations by graphing:
x = -3
x = 2
a) Graph:
b) Is the system inconsistent or consistent? Why?
The system is inconsistent because there is no value of x which satisfies both equations in the system at the same time. This can be seen in the graph since the two equations represent lines which are parallel. Since the two lines are parallel, they will never intersect. A solution to the system would coincide with the point of intersection. Since there is no intersection, there is no solution to the system.
c) Are they independent of dependent equations? Why?
The two equations are dependent, as each equation can be formed as a linear combination of the other. For instance, the line x = -3 can be obtained from the line x = 2 by subtracting the constant value 5 from every point on the line x = 2.
IV) Solve using elimination/addition method
Multiplying the first equation through by 5, and the second equation through by 10 gives the following equivalent system:
Adding the two equations together will eliminate the variable y:
Substituting this into the first equation in the original system and solving for y gives:
Solution: x = -14/55, y = -58/55
V) Last month Jerry purchases two DVDs and five CDs at Wall-to-Wall Sound for $65. This month he bought four DVDs and three CDs for $81. Find the price of
each DVD and find the price of each CD. [Use either addition or substitution
method.]
Let D represent the price of a DVD, and let C represent the price of a CD.
From Jerry’s first purchase, the following equation can be written:
2D + 5C = 65
From Jerry’s second purchase, the following equation can be written:
4D + 3C = 81
This gives the following system of equations:
2D + 5C = 65
4D + 3C = 81
The prices are the solution to this system of equations.
Using the addition method, and multiplying the first equation by -2 gives:
-4D – 10C = -130
4D + 3C = 81
Adding the two equations together eliminates the variable D, leaving:
-7C = -49
Solving for C then gives:
C = -49 / -7
C = 7
Then, substituting this into the first equation of the original system and solving for D gives:
2D + 5C = 65
2D + 5(7) = 65
2D + 35 = 65
2D = 65 – 35
2D = 30
D = 30 / 2
D = 15
Solution: The price of each DVD is $15, and the price of each CD is $7.
3x + 30x( )+ −5y + 5y( ) = 4 + −18( )
33x + 0y = −14
x = −14 33
3x+30x
()
+-5y+5y
()
=4+-18
()
33x+0y=-14
x=
-14
33
3 5
− 14 33
⎛ ⎝⎜
⎞ ⎠⎟ − y =
4 5
y = 3 5
− 14 33
⎛ ⎝⎜
⎞ ⎠⎟ − 4 5
y = − 14 55
− 4 5
3
5
-
14
33
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-y=
4
5
y=
3
5
-
14
33
æ
è
ç
ö
ø
÷
-
4
5
y=-
14
55
-
4
5
y = − 14 55
− 4 11( ) 5 11( )
y = − 14 55
− 44 55
y = − 58 55
y=-
14
55
-
411
()
511
()
y=-
14
55
-
44
55
y=-
58
55
3 5 x − y =
4 5
3x + y 2 = −
9 5
3
5
x-y=
4
5
3x+
y
2
=-
9
5
3x − 5y = 4
30x + 5y = −18
3x-5y=4
30x+5y=-18
