Question

Solve the following system of equations.

2x + 3y – z = 1
3x + y + 2z = 12
x + 2y – 3z = –5

Answer 1

You can do simultaneous equations few times or use vector,
x=3,y=-1,z=2

Answer 2

Okay, thank you

Answer 3

2x + 3y – z = 1
x = (1- 3y + z )/2..................(1)
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3 (1- 3y + z )/2 + y + 2z = 12

(3(1-3y+z))/(2)+y+2z=12

(3(-3y+z+1))/(2)+y+2z=12

(3(-3y+z+1))/(2)+y*(2)/(2)+2z*(2)/(2)=12

(3(-3y+z+1))/(2)+(2y)/(2)+(4z)/(2)=12

(3(-3y+z+1)+(2y)+(4z))/(2)=12

(-7y+7z+3)/(2)=12

-7y+7z+3=24

-7y=-7z+21

y=z-3.....................(2)
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From eqn 3,

(1- 3(z-3) + z )/2 + 2(z-3) – 3z = –5

(1-3(z-3)+z)/(2)+2(z-3)-3z=-5

(1-3z+9+z)/(2)+2(z-3)-3z=-5

(10-2z)/(2)+2(z-3)-3z=-5

(2(5-z))/(2)+2(z-3)-3z=-5

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

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

5-z+2z-6-3z=-5

-1-2z=-5

-2z=-4

z=2...................(3)
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Sub (3) to (2)
y=-1
Sub (3) and (2) to (1)
x=3
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x=3,y=-1,z=2