# Solve the following system of equations: x+2z=2 2y-z=3 x+2y+4z=-1

## Question:

Solve the following system of equations:

{eq}x+2z=2 \\ 2y-z=3 \\ x+2y+4z=-1 {/eq}

## System of Equations:

A system of three equations in three variables is solved by many ways, namely elimination method, substitution method, Cramer's Rule, Gauss Elimination method. Here, we substituted the value of x and y in terms of y using first and second equation and substituted them into the third equation, yielding the value of z. The value of z is then used to obtain the value of x and y.

Given the system

{eq}\displaystyle x+2z=2 \\ 2y-z=3 \\ x+2y+4z=-1 {/eq}

{eq}\displaystyle x+2z=2 \Rightarrow x=2-2z \\ \displaystyle 2y-z=3 \Rightarrow y=\frac{z+3}{2} \\ \displaystyle x+2y+4z=-1\\ \displaystyle \Rightarrow (2-2z)+(z+3)+4z=-1\\ \displaystyle \Rightarrow 3z+5=-1\\ \displaystyle \Rightarrow 3z=-6\\ \displaystyle \Rightarrow z=-2\\ \displaystyle \Rightarrow x=2-2(-2)=6 \\ \displaystyle And \ y=\frac{-2+3}{2}=\frac{1}{2} \\ {/eq}

Therefore, the solution to the system is {eq}\displaystyle \color{blue}{x=6, \ y=\frac 1 2, \ z=-2}. {/eq}