Solve the following system of equations:

$x-y+z=4$$x-2y-2z=9$$2x+y+3z=1$

Given:

The given system of equations is:

$x-y+z=4$

$x-2y-2z=9$

$2x+y+3z=1$

To do:

We have to solve the given system of equations.

Solution:

The given system of equations are:

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

$x-2y-2z=9$....(ii)

$2x+y+3z=1$.....(iii)

From equation (i), we get,

$z=-x+y+4$

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

$x-2y-2(-x+y+4)=9$

$x-2y+2x-2y-8=9$

$3x-4y-8-9=0$

$3x-4y-17=0$.....(iv)

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

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

$2x+y-3x+3y+12=1$

$-x+4y+12-1=0$

$x=4y+11$....(v)

Substituting equation (v) in equation (iv), we get,

$3(4y+11)-4y-17=0$

$12y+33-4y-17=0$

$8y+16=0$

$8y=-16$

$y=\frac{-16}{8}$

$y=-2$

Using $y=-2$ in equation (v), we get,

$x=4(-2)+11$

$x=-8+11$

$x=3$

Substituting the values of $x=3$ and $y=-2$ in $z=-x+y+4$, we get,

$z=-3+(-2)+4$

$z=-5+4$

$z=-1$

Therefore, the solution of the given system of equations is $x=3, y=-2$ and $z=-1$.  

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Updated on: 10-Oct-2022

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