Question

Solve the following system of equations.
2x – 3y + z = –1
3x + 2y + 2z = –1
x – y – 3z = –4

Answer 1

if you use elimination
6x-2y=-6
solve for a variable subtract 6x
-2y=-6x-6
divide by -2
y=3x+3
substitute into the second equation
3x+2y+2z=-1
3x+2(3x+3)+2z=-1
2x+6x+6+2z=-1
combine like terms
8x+6+2z=-1
subtract 6
8x+2z=-7
subtract 8x
2z=8x+7
divide by 2
z=4x+(7/2)
substitute the y and z solutions into the third equation
x – y – 3z = –4
x-(3x+3)-3(4x+(7/2)=-4
distribute
x-3x-3-12x-(21/2)=-4
combine like terms
-14x-(27/2)=-4
add 27/2
-14x=(19/2)
divide by -14
x=(-19/28)
substitute x in for the y and z equations
y=3(-19/28)+3
PEMDAS
y=27/28
z=4(-19/28)+(7/2)
PEMDAS
z=11/14
so your coordinate triple is \[\left( \frac{ -19 }{ 28 },\frac{ 27 }{ 28 },\frac{ 11 }{ 14 } \right)\]

Answer 2

sorry it's quite long there is a lot of work for this kind of thing