Solve the following system of equations:

$x-y+z=4$$x+y+z=2$$2x+y-3z=0$

Given:

The given system of equations is:

$x-y+z=4$

$x+y+z=2$

$2x+y-3z=0$

To do:

We have to solve the given system of equations.

Solution:

The given system of equations are:

$x-y+z=4$....(i)

$x+y+z=2$....(ii)

$2x+y-3z=0$.....(iii)

From equation (i), we get,

$z=-x+y+4$

Substituting the value of $z=-x+y+4$ in equation (ii), we get,

$x+y+(-x+y+4)=2$

$x+y-x+y+4=2$

$2y+4=2$

$2y=2-4$

$2y=-2$

$y=\frac{-2}{2}$

$y=-1$.....(iv)

Substituting the value of $z=-x+y+4$ and $y=-1$ in equation (iii), we get,

$2x+(-1)-3(-x-1+4)=0$

$2x-1+3x-9=0$

$5x-10=0$

$5x=10$

$x=\frac{10}{5}$

$x=2$....(v)

Substituting $x=2$ and $y=-1$ in $z=-x+y+4$, we get,

$z=-2+(-1)+4$

$z=4-3$

$z=1$

Therefore, the solution of the given system of equations is $x=2, y=-1$ and $z=1$.  

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Updated on: 10-Oct-2022

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