Question 56165: solve the system:
-32+3z=4(x-2y)
4(x-2y-z)=-36
-2(2x+y)+2z=-12
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! A tedious looking problem which is no doubt created to torment young minds, however, on closer inspection we can see that it can be unravelled without too much difficulty.
:
Get each equation in the ax + by + cz = d format:
Eq1:
32 + 3z = 4(x-2y)
32 + 3z = 4x - 8y
-4x + 8y + 3z = -32
:
Eq2:
4(x-2y-z) = -36
4x - 8y - 4z = -36
:
:
Eq3:
-2(2x+y) + 2z = -12
-4x - 2y + 2z = -12
:
;
Let's just deal with the 1st two equations; notice if we add them we will eliminate x and y, leaving us with the solution to z:
:
-4x + 8y + 3z = -32
4x - 8y - 4z = - 36
-------------------- add
0x + 0y - 1z = -68
z = -68/-1
z = +68
:
Take eq 2 & 3, substitute 68 for z giving us two equations with two unknowns:
Eq2: 4x - 8y - 4z = -36 > > 4x - 8y - 4(68) = -36 > > 4x - 8y - 272 = -36
Eq3 -4x - 2y + 2z = -12 > > -4x - 2y + 2(68) = -12 > > -4x -2y + 136 = -12
:
Combining the numbers:
4x - 8y = -36 + 272
-4x -2y = -12 - 136
:
We have two managable equations, adding them eliminates x, giving the y solution
4x - 8y = + 236
-4x - 2y = - 148
---------------- add
0x - 10y = + 88
y = +88/-10
y = - 8.8
:
Now we have y = -8.8 and z = + 68, almost done, let's substitute for y & z in the 3rd equation
-4x - 2y + 2z = -12
-4x - 2(-8.8) + 2(68) = -12
:
-4x + 17.6 + 136 = -12
-4x = -12 - 17.6 - 136
-4x = -165.6
x = -165.6/-4
x = +41.4
:
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