Hello,

This is the last of questions I have for my practice assignment. Again I will put what I got. Thanks for your help.

1.) Solve the system using substitution. Write answer as an ordered pair.
4x+2y=8
-5x-2=y

A.) First equation (-2,8)
B.) Second equation (-2,8)

I put A

2.) Solve the system using elimination. If the system is linearly dependent, state the general solution in terms of a parameter.

4x-2y+2z=2
-x+y-3z=-4
4x-3y+7z=9

A.) (-1,-7,0)
B.) (3,3,2)
C.) (2z-3, 5z-7, z)
D.) no solution, inconsistent

I got D.

3. Solve using the method of your choice.
y-2^x^2+2x=0
y=4^x+8

A) (4,2^24)
B) (-4, 2^8)
C) (4, 2^24) (-4, 2^8)
D) (4, 2^24) (-4, -2^8)

I got B

1. From substitution we get 4-2x= -5x-2

So, 3x = -6, x=-2, so y = 8: (-2,8)

2. Should be D

For the first question, to solve the system using substitution, we need to isolate one variable in one of the equations and substitute it into the other equation. Let's solve it step by step:

1.) Solve the first equation for x:
4x + 2y = 8
Subtract 2y from both sides:
4x = 8 - 2y
Divide both sides by 4:
x = 2 - 0.5y

2.) Substitute x into the second equation:
-5x - 2 = y
Replace x with 2 - 0.5y:
-5(2 - 0.5y) - 2 = y
Distribute the -5:
-10 + 2.5y - 2 = y
Combine like terms:
0.5y - 12 = y
Subtract 0.5y from both sides:
-12 = 0.5y
Divide both sides by 0.5:
-24 = y

So the ordered pair solution is (x, y) = (-2, -24). None of the given options are correct, so the correct answer should be "None of the above".

For the second question, to solve the system using elimination, we need to eliminate one variable by multiplying one or both of the equations by appropriate constants so that the coefficients of one variable are equal (or additive inverses).

1.) Multiply the second equation by 4, and the third equation by -1:
4(x - y + 3z) = -4(-1)
-4(4x - 3y + 7z) = -4(9)
This will give us:
4x - 4y + 12z = 4
-16x + 12y - 28z = -36

2.) Add these two equations together:
(4x - 4y + 12z) + (-16x + 12y - 28z) = 4 + (-36)
Combine like terms:
-12x - 16z = -32

This equation only has two variables, x and z, and no y. Therefore, it represents a plane in a three-dimensional space, and we can't determine a unique solution for x, y, and z. The system is linearly dependent, and the general solution can be expressed in terms of a parameter.

So the correct answer is C) (2z-3, 5z-7, z), which represents the general solution in terms of the parameter z.

For the third question, to solve the system using the method of your choice, we can substitute the value of y from the second equation into the first equation:

1.) Substitute y = 4^x + 8 into the first equation:
y - 2^(x^2 + 2x) = 0
4^x + 8 - 2^(x^2 + 2x) = 0

2.) Simplify and rearrange the equation:
2^(2x) - 2^(x^2 + 2x) = -8
2^(2x) = 2^(x^2 + 2x - 3)
Since the bases (2) are the same, we can equate the exponents:
2x = x^2 + 2x - 3

3.) Rearrange and solve the equation for x:
0 = x^2 - 3
x^2 = 3
x = ±√3

4.) Substitute x back into one of the original equations to solve for y:
y = 4^x + 8
For x = √3:
y = 4^(√3) + 8 ≈ 17.31
For x = -√3:
y = 4^(-√3) + 8 ≈ -5.52

So the solutions for the system are (x, y) = (√3, 17.31) and (x, y) = (-√3, -5.52). None of the given options match this, so the correct answer should be "None of the above".