What is the answer the to this?

Question

Solve the sytem of linear equations?

3x-2y+z=4
x+3y-4z=-3
2x-3y+5z=7


Im not sure how to do this

anyhelp?

Answer 1

12x - 8y + 4z = 16
x + 3y - 4z = - 3--------ADD

13x - 5y = 13

- 15x + 10y - 5z = - 20
2x - 3y + 5z = 7-------ADD

- 13x + 7y = - 13
13x - 5y = 13-----ADD
2y = 0
y = 0
x = 1

3 - 0 + z = 4
z = 1

x = 1 , y = 0 , z = 1

Answer 2

Isolate 1 variable at a time and plug it into the other given equations. In the above question, we can arrange:
3x - 2y + z = 4 into z = 4 - 3x + 2y
therefore:
x + 3y - 4z = -3 will be x + 3y - 4(4 - 3x + 2y) = -3
which equals to:
13x - 5y = 13
next we do the same with the last equation:
2x - 3y + 5z = 7 will be 2x - 3y + 5(4 - 3x + 2y) = 7
which equals to:
-13x + 7y = -13
Now we are left with 2 equations:
13x - 5y = 13 and -13x + 7y = -13.

We can further manipulate -13x + 7y = -13 into 13x - 7y = 13 which can again be changed into 13x = 13 + 7y. Now we plug this into our other equation to get 13 + 7y - 5y = 13. From this, we know y is 0.

Now we can use this value to figure out that 13x - 5(0) = 13, meaning that x = 1.

Knowing the values of x = 1 and y = 0, we can plug the values into the first original equation to get: 3(1) - 2(0) + z = 4, therefore 3 - 0 + z = 4, therefore z = 1.

Therefore, x = 1, y = 0, z = 1.

Answer 3

3x-2y+z=4
x+3y-4z=-3
2x-3y+5z=7

Add to 2nd and 3rd equations
x+3y-4z=-3
2x-3y+5z=7
_________
3x +0y + z = 4
or
3x + z = 4

SUBTRACT this equation from the first equation
3x-2y+z=4
-(3x + z) = -(4)
-2y = 0
y = 0

then you have x = 1 and z = 1 by substituting the value for y into the other equations

ANSWER
y = 0
x = 1
z = 1

these values check in all three original equations

Answer 4

use substitution method