Question

Solve for x, y and z in the given system of equations using Elimination :

$x=3z-52x+2z=y+167x-5z=3y+19$

• A.
  $4,-2and3$
• B.
  $4,2and-3$
• C.
  $5,4and4$
• D.
  $1,-2and4$

Answer 1

The correct option is A

$4,-2and3$
$x-3z+5=\mathrm{0...}\left(1\right)2x-2z-y-16=\mathrm{0...}\left(2\right)7x-5z-3y-19=\mathrm{0...}\left(3\right)$

Multiply by 3 on both sides in (2),

$6x+6z-3y-48=0$
Subtracting above equation from (3), we get

$7x-5z-3y-19-(6x+6z-3y)=48\Rightarrow x-11z+29=0-(-x+11z-29=0\therefore -x+11z-29=\mathrm{0...}(4)$

Adding (1)and (4), we get :

$x-3z+5+(-x+11z-29)=0\Rightarrow 8z-24=0\Rightarrow 8z=24\Rightarrow z=\frac{24}{8}\therefore z=3$

Substituting z=3 in 1, we get :

$x-3z+5=0\Rightarrow x-3\left(3\right)+5=0\Rightarrow x-9+5=0\Rightarrow x=4$

Substituting x=4,z=3 in (2), we get :

$2x+2z-y-16=0\Rightarrow 2\left(4\right)+2\left(3\right)-y-16=0\Rightarrow 8+6-y-16=0\Rightarrow 14-y-16=0\Rightarrow -2-y=0\Rightarrow -y=2\therefore y=-2$

So, option a is correct.