Chapter 13, Problem 6STP

To determine

To find:Solution of the system of equations.

Answer to Problem 6STP

  $\left(x,y,z\right)=\left(2,5,3\right)$

Explanation of Solution

Given information:

  $\begin{array}{l}x-y+z=0\\ -5x+3y-2z=-1\\ 2x-y+4z=11\end{array}$

  $x-y+z=0$ ------(1)

  $-5x+3y-2z=-1$ --------(2)

  $2x-y+4z=11$ ---------(3)

On Subtracting equation-(3) with equation-(1)

We get

  $\begin{array}{l}2x-y+4z-\left(x-y+z\right)=11-0\\ \Rightarrow 2x-y+4z-x-\left(-y\right)-z=11\\ \Rightarrow 2x-y+4z-x+y-z=11\\ \Rightarrow x+3z=11\end{array}$

  $x+3z=11$ --------(4)

Multiplying equation-(1) with $3$ and subtracting with equation-(2)

We get

  $\begin{array}{l}\Rightarrow 3\left(x-y+z\right)+\left(-5x+3y-2z\right)=3\left(0\right)+\left(-1\right)\\ \Rightarrow 3x-3y+3z-5x+3y-2z=0-1\\ \Rightarrow 3x-5x-3y+3y+3z-2z=-1\\ \Rightarrow -2x+z=-1\\ \Rightarrow 2x-z=1\end{array}$

  $2x-z=1$ -------(5)

Multiplying equation-(4) with $2$ and subtracting with equation-(5)

We get

  $\begin{array}{l}2\left(x+3z\right)-\left(2x-z\right)=2\left(11\right)-1\\ \Rightarrow 2x+2\left(3z\right)-2x-\left(-z\right)=22-1\\ \Rightarrow 2x+6z-2x+z=21\\ \Rightarrow 6z+z=21\\ \Rightarrow 7z=21\\ \Rightarrow \frac{7z}{7}=\frac{21}{7}\\ \Rightarrow z=3\end{array}$

Substitute value of $z$ in equation-(5)

We get

  $\begin{array}{l}2x-z=1\\ \Rightarrow 2x-3=1\\ \Rightarrow 2x-3+3=1+3\\ \Rightarrow 2x=4\\ \Rightarrow \frac{2x}{2}=\frac{4}{2}\\ \Rightarrow x=2\end{array}$

Substitute values of $x,z$ in equation-(1)

We get

  $\begin{array}{l}x-y+z=0\\ \Rightarrow 2-y+3=0\\ \Rightarrow 5-y=0\\ \Rightarrow 5-y+y=0+y\\ \Rightarrow y=5\end{array}$

Therefore,

Solution set is

  $\left(x,y,z\right)=\left(2,5,3\right)$