Question

Please help
What is the solution to the system of equations?

-4x+4y-2z=-8

-3x-y+4z=0

2x-2y+3z=-4

a.(1, –7, –2)
b.(–3, –7, –4)
c.(–3, 0, 5)
d.(3, 5, –4)

Answer 1

Answer is b

b.(–3, –7, –4)

Answer 2

Can you explain how you got that answer please

Answer 3

@guruguru

Answer 4

-4x+4y-2z=-8
-3x-y+4z=0 X -2 Multiply by -2 to match with third equation
2x-2y+3z=-4


-4x+4y-2z=-8
______________________
6x+2y-8z=0
2x-2y+3z=-4
______________________
8x -5z =-4

Add second and third equation, this will eliminate y variable from second and third equation

You will get
8x -5z =-4

Now you can multiply

-3x-y+4z=0 X 4 Multiply by 4 to match with First equation

The reason we want to do is to eliminate Y variable from all three equations

after multiplication, you need to add equation first and second, this way it will cross out y variable from both first and second equations

-4x + 4y - 2z =-8
-12x -4y +16z =0
_______________________
-16x +14z = -8
_______________________


Now you get two equations after eliminating all three equations.

Equation1 = 8x -5z = -4
Equation2 = -16x +14z = -8

We don't have y variable now, we need to eliminate x. After eliminating x we will get z value.

To eliminate x, we need to multiply first equation with 2

8x -5z = -4 X 2
-16x +14z = -8

You will get

16x - 10z = -8
-16x +14z = -8
______________________
4z = -16 So z = -4
______________________


Now plug z value in the second equation we got, its your choice either first or second equation. I use second equation to get x value.

-16x +14z = -8

-16x +14(-4) = -8
-16x - 56 =-8
-16x = -8+56
-16x = 48
x = -3

So you get x and z value. Plug both values in any of the three equations in the question to get the y value.

so i use equation 1

-4x+4y-2z=-8
-4(-3)+4y-2(-4) = -8
12+4y+8 = -8
20+4y = -8
4y = -8-20
4y = -28
y = -7


This is how you get the result
X = -3
Y = -7
Z = -4

Answer 5

@jkbo its easy, just need to practice little bit.

Answer 6

thank you very much

Answer 7

no worries friend !