Question

Solve the following system of linear equations in three variables.
$x+y+z=52x-y+z=9x-2y+3z=16$

• A. $x=18y=\frac{-19}{3}z=-4$
• B. $x=2y=-1z=4$
• C. $x=-4y=18z=-\frac{19}{3}$
• D. $x=-\frac{19}{3}y=18z=-4$

Answer 1

The correct option is B $x=2y=-1z=4$

$x+y+z=5-----\left(1\right)2x-y+z=9-----\left(2\right)x-2y+3z=16----\left(3\right)$

$eqn\left(1\right)+eqn\left(2\right)gives$
x + y +z = 5
2x-y +z = 9
______________
3x +2z = 14 --------------(4)
$eqn\left(1\right)\times 2+eqn\left(3\right)gives$
2x + 2y +2z = 10
x-2y +3z = 16
______________
3x +5z = 26 --------------(5)
$eqn\left(4\right)-eqn\left(5\right)gives$
3x +2z = 14
3x +5z = 26
- - -
______________
-3z = -12
z= 4 ------------(6)
Substituting eqn(6) in eqn(4) we get,
$3x+2\times 4=5$
$3x+8=14$
$x=2$ --------------(7)
Substituting eqn(6) and eqn(7) in eqn(1) we get,
$2+y+4=5$
$y=5-6$
$y=-1$ -------------(8)
Verification
Substituting the values of x,y and z in eqn(2) we get,
$2\times 2-(-1)+4=4+1+4$
$=9$
$RHS$