Question

Note: Make sure to show all necessary steps on your scratch paper/work document. What is the value of y in the system of equations below? 3x+y+2z=−10 x+5y−2z=−12 −2x+3y=−1

Answer 1

State the Objective Function (using the variable S for score) that could be used to maximize the exam score in the linear programming problem given.(1 point) Responses S=3x+5y cap s is equal to 3 x plus 5 y S=5x+3y cap s is equal to 5 x plus 3 y S=2x+6y cap s is equal to 2 x plus 6 y S=6x+2y

Answer 2

To solve this system of equations, we can use the method of elimination or substitution.

Let's use the method of elimination:
1) Multiply the second equation by 3 to create a cancellation with the third equation.
Original second equation: x + 5y - 2z = -12
Multiplying by 3: 3(x + 5y - 2z) = 3(-12)
Simplifying: 3x + 15y - 6z = -36

2) Now we can add the modified second equation to the third equation to eliminate x.
Modified second equation: 3x + 15y - 6z = -36
Third equation: -2x + 3y = -1
Adding the equations: (3x + 15y - 6z) + (-2x + 3y) = -36 + (-1)
Simplifying: x + 18y - 6z = -37

3) Rewrite the first equation with the variables in the same order as the previous equation.
Original first equation: 3x + y + 2z = -10

4) Now we have two equations:
x + 18y - 6z = -37
3x + y + 2z = -10

We can eliminate x by multiplying the first equation by -3 and adding it to the second equation:
-3(x + 18y - 6z) + (3x + y + 2z) = -3(-37) + (-10)
-3x - 54y + 18z + 3x + y + 2z = 111 + (-10)
Simplifying: -53y + 20z = 101

5) Now we have these equations:
-53y + 20z = 101
3x + y + 2z = -10

6) We can eliminate y by multiplying the first equation by 3 and adding it to the second equation:
3(-53y + 20z) + (3x + y + 2z) = 3(101) + (-10)
-159y + 60z + 3x + y + 2z = 303 + (-10)
Simplifying: -158y + 62z + 3x = 293

7) Now we have these equations:
-158y + 62z + 3x = 293
3x + y + 2z = -10

8) Equate the coefficients of x in both equations:
3x = 3x
Therefore, the value of x is irrelevant to solving for y and z, so we can ignore x.

9) Now we have these simplified equations:
-158y + 62z = 293
y + 2z = -10

10) We can solve the second equation for y:
y = -2z - 10

11) Substitute the value of y in the first equation:
-158(-2z - 10) + 62z = 293
316z + 1580 + 62z = 293
378z + 1580 = 293
378z = 293 - 1580
378z = -1287
z = -1287/378
z = -3.4

12) Now substitute the value of z in the second equation to find y:
y + 2(-3.4) = -10
y - 6.8 = -10
y = -10 + 6.8
y = -3.2

So, in the given system of equations, the value of y is -3.2.

Answer 3

The objective function that could be used to maximize the exam score in the linear programming problem is:

S = 3x + 5y

where S represents the score, and x and y are variables representing different factors that contribute to the exam score.