3x+2y-z+5

-2x-y+3z=2

2x+2y+4z=14

solve linear equation with 3 varables using the addition/elimination method.

x+y+z=3,   2x-5y-3z=-7,    x-z=-2
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1 Answer

(a)

(1) 3x+2y-z=5

(2) -2x-y+3z=2

(3) 2x+2y+4z=14

Double (2): -4x-2y+6z=4 and add to (1): -x+5z=9. Now we need to eliminate y using (3) and one of the others:

Multiply the original (3) by -1: -2x-2y-4z=-14 and add to (1):

x-5z=-9. 

We have two equations, but they are the same! This means that there is no unique solution, but we can establish some relationships.

x=5z-9 relates x and z. Equation (3) can be simplified: x+y+2z=7 then substitute for x: 5z-9+y+2z=7, y+7z=16, so y=16-7z.

We have x=5z-9 and y=16-7z. This means we can substitute any value for z and then calculate x and y using these relationships.

EXAMPLE: z=1, so x=-4 and y=9. Now test these in the original equations:

(1) -12+18-1=5✔️

(2) 8-9+3=2✔️

(3) -8+18+4=14✔️

If we now put z=0, then x=-9 and y=16. Test again:

(1) -27+32=5✔️

(2) 18-16=2✔️

(3) -18+32=14✔️

It's clear that there are many solutions but there are consistent relationships: x=5z-9 and y=16-7z.

(b)

(1) x+y+z=3

(2) 2x-5y-3z=-7

(3) x-z=-2, so x=z-2, we can substitute for x in the other two equations:

(4) z-2+y+z=3, y+2z=5

(5) 2z-4-5y-3z=-7, -5y-z=-3, doubled: -10y-2z=-6.

Add (5) to (4): -9y=-1, y=1/9.

Substitute for x and y in (1) 2z-4-5/9-3z=-7, 22/9=z, z=22/9; x=z-2=22/9-2=4/9.

These values fit all three equations for a unique solution: x=4/9, y=1/9, z=22/9.

by Top Rated User (1.2m points)

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