Question

Solve the following systems of equations:
$x-y+z=4$
$x+y+z=2$
$2x+y-3z=0$

Answer 1

$x-y+z=4⟶\left(i\right)x+y+z=2⟶\left(ii\right)2x+y-3z=0⟶\left(iii\right)$

Solving $\left(iii\right),$ $y=-2x+3z$

Substituting this value in $\left(i\right)$ and $\left(ii\right),$

$\left(i\right)\Rightarrow x-y+z=4\Rightarrow x-(-2x+3z)+z=4\Rightarrow 3x-2z=4⟶\left(iv\right)$

$\left(ii\right)\Rightarrow x+(-2x+3z)+z=2\Rightarrow -x+4z=2⟶\left(v\right)$

Now by solving $\left(iv\right)$ and $\left(v\right),$

$3x-2z=4\times 1\Rightarrow 3x-2z=4⟶\left(vi\right)$

$-x+4z=2\times 3\Rightarrow -3x+12z=6⟶\left(vii\right)$

Adding $\left(vi\right)$ and $\left(vii\right),$

$10z=10$

$z=1$

By substituting the value of $z$ in equation $\left(iv\right),$

$3x-2z=4$

$\Rightarrow 3x-2=4$

$\Rightarrow x=2$

Now by substituting the value of $x$ and $z$ in $\left(i\right),$

$x-y+z=4$

$\Rightarrow 2-y+1=4$

$\Rightarrow -y=1$

$\Rightarrow y=-1$

$\therefore (x,y,z)=(2,-1,1)$