Question

Solve: $x-y+z=6$; $x-2y-2z=5$ and $2x+y-3z=1$

• A. $x=3,y=-2,z=1$
• B. $x=-2,y=1,z=8$
• C. $x=0,y=7,z=-9$
• D. $x=1y=-,z=-1$

Answer 1

The correct option is A $x=3,y=-2,z=1$

$x-y+z=6$ ......(i)
$x-2y-2z=5$ .....(ii)
$2x+y-3z=1$ .....(iii)

Subtract equations (ii) from (i), we get
$y+3z=1$ ......(iv)

Multiply equation (ii) by $2$ and subtract from equation (iii),
$5y+z=-9$ .....(v)

Multiply equation (iv) by $5$
$5y+15z=5$
$5y+z=-9$
-----------------
$14z=14$
$\therefore z=1$

Substitute $z=1$ in equation (iv),
$y+3\left(1\right)=1$
$\therefore y=-2$

Substitute $y=-2$ and $z=1$ in equation (i),
$x-(-2)+1=6$

$\therefore x=3$
So, $x=3,y=-2,z=1$