2z-3y+z=-19

Question

5x+y-z=-7
-x+6y-z=35
What is the solution to the system of equations?

Answer 1

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

First, let's use elimination by adding the second and third equations to eliminate the "z" term:

(5x + y - z) + (-x + 6y - z) = -7 + 35
4x + 7y - 2z = 28

Now, let's add this new equation and the first equation:

(4x + 7y - 2z) + (2z - 3y + z) = 28 - 19
4x + 7y = 9

Next, let's use substitution to solve for one variable. Let's solve the first equation for "z":

2z - 3y + z = -19
3z - 3y = -19
3z = 3y - 19
z = (3y - 19) / 3

Now, let's substitute this value of "z" in the second equation:

5x + y - [(3y - 19) / 3] = -7
5x + y - (3y - 19) / 3 = -7
15x + 3y - 3y + 19 = -21
15x + 19 = -21
15x = -21 - 19
15x = -40
x = -40 / 15
x = -8/3

Now that we have found the values of x and z, let's substitute the values in the third equation:

-(-8/3) + 6y - [(3y - 19) / 3] = 35
8/3 + 6y - (3y - 19) / 3 = 35
8/3 + (18y - 19) / 3 = 35
(8 + 18y - 19) / 3 = 35
18y - 11 = 105
18y = 105 + 11
18y = 116
y = 116 / 18
y = 58/9

Therefore, the solution to the system of equations is x = -8/3, y = 58/9, and z = (3y - 19) / 3.

Answer 2

I suspect a typo -- the first equation is most likely
2x-3y+z = -19

Secondly, adding the 2nd and third equations does not eliminate z, since
(5x + y - z) + (-x + 6y - z) = 4x+7y-2z

In any case solving the system of equations correctly yields a solution of (-2,6,3)

Answer 3

Apologies for the confusion. Let's revise the solution using the corrected first equation and solve the system of equations.

The revised system of equations is:
2x - 3y + z = -19
5x + y - z = -7
-x + 6y - z = 35

To eliminate the z variable, we can add the first and second equations:
(2x - 3y + z) + (5x + y - z) = -19 + (-7)
7x - 2y = -26

Now, let's add this new equation to the third equation:
(7x - 2y) + (-x + 6y - z) = -26 + 35
6x + 4y - z = 9

We now have two equations:
7x - 2y = -26
6x + 4y - z = 9

Let's solve the first equation for x:
7x = -26 + 2y
x = (-26 + 2y) / 7

Now, substitute this value of x into the second equation:
6((-26 + 2y) / 7) + 4y - z = 9
(-156 + 12y) / 7 + 4y - z = 9
-156 + 12y + 28y - 7z = 63
40y - 7z = 219

Now we can solve the equation 7x - 2y = -26 for x and z:
7((-26 + 2y) / 7) - 2y = -26
-26 + 2y - 2y = -26
-26 = -26 (True)

Therefore, x can have any value.

Substitute x = 0 into the equation 40y - 7z = 219:
40y - 7z = 219
40y = 7z + 219
y = (7z + 219) / 40

Since x can have any value, we can choose x = 0. Let's substitute x = 0 into the second equation:
5(0) + y - z = -7
y - z = -7
y = z - 7

Now we can express y and z relative to a parameter t:
y = z - 7
y = (7z + 219) / 40
(7z + 219) / 40 = z - 7
7z + 219 = 40z - 280
33z = 499
z = 499 / 33
z = 15.12

Substitute z = 15.12 back into y = z - 7:
y = z - 7
y = 15.12 - 7
y = 8.12

Therefore, the solution to the system of equations is x = 0, y = 8.12, and z = 15.12