solve by the method of reduction?

Question

x+y-z=7 2x-3y-2z=4 x-y-5z=23

Answer 1

(1) x+y=7+z,

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

(3) x-y=23+5z, 2(3) is 2x-2y=46+10z

(4)=(1)+(3)=2x=30+6z, x=15+3z;

(5)=2(3)-(2)=-2y+3y=46-4+10z-2z, y=42+8z.

So we now have x and y in terms of z in equations (4) and (5).

This means we can take any of the equations and substitute for x and y to find z.

Take (1), for example:

15+3z+42+8z=7+z,

(3+8-1)z=7-15-42=-50, so z=-50/10=-5.

Now we have z we can find x=15+3z=15-15=0; y=42+8z=42-40=2.

SOLUTION: x=0, y=2, z=-5. All equations are satisfied.