Question

Solve the system by elimination

-2x+2y+3z=0
-2x-y+z=-3
2x+3y+3z+5


Solve the system by substitution.

-x-y-z=-8
-4x+4y+5z=7
2x+2z=4

Answer 1

Explanation:

Given system is-

-2x+2y+3z=0 -------------(1)

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

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

From (1)-(2) we get

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

Again from (2)+(3),

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

Now multiplying equation (4) by -2

-6y-4z=-6 --------------(6)

Adding (5) and (6)

-4y=-4

dividing by -4

y=-4 / -4

=1

So, y=1 ,

using y=1 in (5)

2(1)+4z=2

or, 2+4z=2

or, 4z=2-2

or, 4z=0

or, z=0

using y=1 and z=0 in (3)

2x+3(1)+3(0)=5

or, 2x+3+0=5

or, 2x=5-3

or, 2x=2

or, x=2/2

=1

So, the solution is-

(x,y,z)=(1 , 1 , 0)