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Can somebody please walked through this, I'm so confused and I have a test in 6 hours...

Can Somebody Please Walked Through This Im So Confused And I Have A Test In 6 Hours class=

Sagot :

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Answer:

Q3: x = 4, y = 4, z = 4

Q4: x = 6, y = 0, z = -4

Step-by-step explanation:

Question 3: Simultaneous equations requires us to solve for x, y and z.

Since all three equations have a z in them, I will first solve for z.

Substitute in the first and third equation into the second equation.

First equation: x = 5z - 16

Second Equation: -4x + 4y - 5z = -20

Third equation: y = -z + 8

Substituting in x = 5z - 16 and y = -z + 8 for the x and y in the second equation.

-4(5z - 16) + 4(-z + 8) - 5z = -20

Expand

-20z + 64   - 4z + 32   - 5z = -20

Simplify and solve for z by putting all the numbers on one side and all the z's on the other side of the equals

-20z - 4z - 5z = -20 - 32 - 64

-29z = -116

z = -116/-29

z = 4

Substitute in this z value into the first and last equation and then solve for x and y

x = 5z - 16

x = 5(4) - 16

x = 20 - 16

x = 4

And

y = -z + 8

y = -(4) + 8

y = 4 (Its just a coincidence that they all equal to 4, I promise)

Question 5: A little bit harder of a question. Since the first and second equation both only have y and z, we can solve it using the elimination method.

Rearrange them so that the letters are on one side and numbers on the other side.

First equation: y + 6z = -24

Second equation: z + 2y = -4

I will choose to eliminate the y (You can choose either or)

Multiply the first equation by 2

2(y + 6z = -24)

2y + 12z = -48

Now that 2y is in both equations, we can minus one equation from the other to eliminate the y (I will minus the second from the first)

First Eq: 2y + 12z = -48

Second Eq: z + 2y = -4

2y - 2y = 0y

12z - z = 11z

-48 - (-4) = -44

Type these answers into a new equation

0y + 11z = - 44

Since y is 0, ignore it. Solve for z

11z = -44

z = -44/11

z = - 4

Substitute our z into either the first or second equation and solve for y (It doesnt matter which one you choose, I just did the second equation)

z + 2y = -4

(-4) + 2y = -4

2y = -4 + 4

2y = 0

y = 0

Substitute in our y and z values into the third equation and solve for x

-6x - 6y - 6z = -12

-6x - 6(0) - 6(-4) = -12

-6x - 0 + 24 = -12

-6x = -12 - 24

-6x = -36

x = -36/-6

x = 6

djtwinx017

Answer:

x = 4, y = 4, z = 4

Step-by-step explanation:

Given the following systems of linear equations:

Equation 1:    x = 5z - 16

Equation 2 :  -4x + 4y - 5z = -20

Equation 3:    y = -z + 8

Using the substitution method, we could either use the value for x in Equation 1, or the value of y in Equation 3 to substitute in the other given equations.

 

Step 1

Let's use Equation 3, and substitute the value of y = -z + 8 into Equation 2:

Equation 3:    y = -z + 8

Equation 2 :  -4x + 4y - 5z = -20

-4x + 4(-z + 8) - 5z = -20

-4x - 4z + 32 - 5z = -20

-4x - 9z + 32 = -20

Step 2:

Using Equation 1, substitute the value of x = 5z - 16 into the previous step:

Equation 1:    x = 5z - 16 into:

-4(5z - 16) - 9z + 32 = -20

-20z + 64 - 9z + 32 = -20

-29z + 96 = -20

Subtract 96 from both sides:

-29z + 96 - 96 = -20 - 96

-29z = -116

Divide both sides by -29:

[tex]\frac{-29z}{-29} = \frac{-116}{-29}[/tex]

z = 4

Step 3:

Substitute the value of z = 4 into Equation 3:

Equation 3:    y = -z + 8

y = -(4) + 8

y = 4

Step 4

Substitute the values of z into Equation 1 to solve for x:

Equation 1:    x = 5z - 16

x = 5(4) - 16

x = 20 - 16

x = 4

Therefore, x = 4, y = 4, and z = 4.

Step 5

Substitute the values for x, y, and z into the given system to verify that their values are the solutions.

x = 4, y = 4, z = 4

Equation 1:    x = 5z - 16

4 = 5(4) - 16

4 = 20 - 16

4 = 4 (True statement).

Equation 2 :  -4x + 4y - 5z = -20

-4(4) + 4(4) - 5(4) = -20

-16 + 16 - 20 = - 20

0 - 20 = -20

-20 = -20 (True statement).

Equation 3:    y = -z + 8

y = -z + 8

4 = -(4) + 8

4 = -4 + 8

4 = 4 (True statement).

Therefore, x = 4, y = 4, and z = 4 are the solutions to the given systems of linear equations.

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