Question 175229
System of equations
-4x+4y-2z=-8
-3x-y+4z=0
2x-2y+3z=-4 
:
There are several ways to solve this, but this one lends itself to the elimination method.
:
Multiply the 3rd equation by 2 and and add to the 1st equation
-4x + 4y - 2z = -8
+4x - 4y + 6z = -8 
---------------------addition eliminates x & y, easy to find z
0x + 0y + 4z = -16
z = {{{(-16)/4}}}
z = -4
:
Substitute -4 for z in the 2nd equation
-3x - y + 4(-4) = 0
-3x - y - 16 = 0
-3x - y = 16
:
Substitute -4 for z in the 3rd equation
2x - 2y +3(-4) = -4
2x - 2y - 12 = -4
2x - 2y = -4 + 12
2x - 2y = +8
:
Using these two equations multiply the 1st, two unknown equation, by -2
Add to the above equation
6x + 2y = -32
2x - 2y = 8
-----------------addition eliminates y, find x
8x + 0y = -24
x = {{{(-24)/8}}}
x = -3
:
Using the 1st original equation substitute -3 for x and -4 for z
-4x + 4y - 2z = -8
-4(-3) + 4y - 2(-4) = -8
+12 + 4y + 8 = -8
4y + 20 = -8
4y = -8 - 20
4y = -28
y = {{{(-28)/4}}}
y = -7
:
:
Check solutions of x=-3; y=-7; z=-4 in the 2nd original equation:
-3x - y + 4z = 0
-3(-3) - (-7) + 4(-4) = 0
+9 + 7 - 16 = 0; confirms our solutions