Question

Solve the system by elimination

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

Answer 1

In point form, (1, 1, 0), or equation form:

`x=1,y=1,z=0`

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How So?

Well,....

Isolate x for -2x + 2y + 3z = 0:

`x=(2y+3z)/(2)`

Substitute
`x=(2y+3z)/(2)`:

`[-2*(2y+3z)/(2)-y+z=-3; 2*(2y+3z)/(2)+3y+3z=5]`

Simplify:

`[-3y-2z=-3; 5y+6z=5]`

Isolate y for -3y - 2z = -3:

`y=-(-3+2z)/(3)`

Substitute y = - -3+2z/3:

`[5(-(-3+2z)/(3))+6z=5`

Simplify:

`[(15+8z)/(3)=5]`

Isolate z for [15+8z/3 = 5]:

For y = - -3+2z/3

Substitute z = 0

`y=-(-3+2 * 0)/(3)`
`=1`

`y=1`

For x = 2y+3z/2

Substitute z = 0, y = 1

`x =(2*1+3*0)/(2)`

Simplify:

`x = 1`

Hence, the solutions to the system of equations are:

`x = 1, y=1, z=0`

Hope this helps!