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
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.
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
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.
To find the value of y in the system of equations, we will use the elimination or substitution method. Let's start by using the elimination method.
Step 1: Multiply the third equation by 3 to make the coefficients of x in both the first and third equations the same:
-6x + 9y = -3
Step 2: Add the first and second equations together to eliminate x:
(3x + y + 2z) + (x + 5y - 2z) = -10 + (-12)
4x + 6y = -22
Step 3: Rewrite the second equation:
4x + 6y = -22
Step 4: Use the third equation (-6x + 9y = -3) and the new equation (4x + 6y = -22) to eliminate y:
(-6x + 9y) - (4x + 6y) = -3 - (-22)
-10x + 3y = 19
Step 5: Solve the resulting equation for x:
-10x + 3y = 19
-10x = 19 - 3y
x = (19 - 3y) / -10
Step 6: Substitute the value of x in terms of y into the equation 4x + 6y = -22:
4((19 - 3y) / -10) + 6y = -22
Step 7: Simplify the equation:
(76 - 12y) / -10 + 6y = -22
Step 8: Multiply the equation through by 10 to get rid of the fraction:
76 - 12y + 60y = -220
Step 9: Combine like terms:
-12y + 60y = -220 - 76
48y = -296
Step 10: Solve for y:
y = -296 / 48
y = -37/6
The value of y in the system of equations is -37/6.
To find the value of y in the given system of equations, we'll solve the system using either substitution or elimination method.
Method 1: Substitution
1. Start with the first equation: 3x + y + 2z = -10.
2. Rearrange the equation to solve for y: y = -3x - 2z - 10.
3. Substitute the value of y in the other equations:
- In the second equation: x + 5(-3x - 2z - 10) - 2z = -12.
- Simplify and solve for x:
x - 15x - 10z - 50 - 2z = -12,
-14x - 12z = 38.
- In the third equation: -2x + 3(-3x - 2z - 10) = -1.
- Simplify and solve for x:
-2x - 9x - 6z - 30 = -1,
-11x - 6z = 29.
4. Now we have a system of two equations:
-14x - 12z = 38,
-11x - 6z = 29.
5. Solve the system of equations for x and z.
- Multiply the first equation by -11 and the second equation by -14 to eliminate x:
154x + 132z = -418,
154x + 84z = -406.
- Subtract the second equation from the first equation to eliminate x:
(154x - 154x) + (132z - 84z) = (-418) - (-406),
48z = -12.
- Divide both sides by 48: z = -12 / 48,
z = -1/4.
6. Substitute the value of z back into one of the original equations to find x:
-11x - 6(-1/4) = 29,
-11x + 3/2 = 29,
-11x = 29 - 3/2,
-11x = 55/2.
Divide both sides by -11: x = (55/2) / (-11),
x = 5/2.
7. Substitute the values of x and z back into the first equation to find y:
3(5/2) + y + 2(-1/4) = -10,
15/2 + y - 1/2 = -10,
y + 14/2 = -10,
y + 7 = -10,
y = -17.
Therefore, the value of y in the system of equations is -17.
Method 2: Elimination
1. Multiply the second equation by 3 and the third equation by 5 to create a new system:
3x + y + 2z = -10,
3x + 15y - 6z = -36,
-10x + 15y = -5.
2. Add the second equation to the third equation to eliminate x:
(3x + 15y - 6z) + (-10x + 15y) = -36 + (-5),
-7x + 30y - 6z = -41.
3. Now we have a new system of equations:
3x + y + 2z = -10,
-7x + 30y - 6z = -41.
4. Multiply the first equation by 7 and subtract it from the second equation to eliminate x:
(-7x + 30y - 6z) - (7x + 7y + 14z) = -41 - (-70),
-7y - 20z = 29.
5. Now we have a system of two equations:
-7y - 20z = 29,
-7x + 30y - 6z = -41.
6. Multiply the first equation by -1 and add it to the second equation to eliminate y:
(-7x + 30y - 6z) + (7y + 20z) = -41 + (-29),
-7x + 27y + 14z = -70.
7. Now we have a new system of equations:
-7x + 27y + 14z = -70,
-7y - 20z = 29.
8. Multiply the second equation by -7 and add it to the first equation to eliminate y:
(-7x + 27y + 14z) + (49y + 140z) = -70 + (-203),
-7x + 76z = -273.
9. Solve the system of equations -7x + 76z = -273 and -7y - 20z = 29 for x and z.
From equation 2: -7y - 20z = 29,
-7y = 29 + 20z,
-7y = 20z + 29.
Divide by -7: y = (20z + 29) / -7,
y = -20z/7 - 29/7.
Substitute this value of y in equation 1: -7x + 76z = -273.
-7x + 76z = -273,
-7x + 76z = -273.
10. Solve the resulting equation for x: -7x = -273 - 76z,
-7x = -273 -76z,
x = (-273 - 76z) / -7,
x = 39 + (76/7)z.
11. Since there are infinitely many solutions for this system, we can let z = 0 to find a particular solution.
Letting z = 0, we have:
x = 39 + (76/7)(0) = 39,
y = -20(0)/7 - 29/7 = -29/7.
12. Therefore, the value of y in the system of equations is -29/7, which can also be expressed as approximately -4.143.
Note: The solution obtained using the elimination method is given as a parameterized form (in terms of z), indicating there are infinitely many solutions. By selecting a specific value for z, we can find a particular solution.