Chapter 9, Problem 4CLT

To determine

To find: Values of x,y and z.

Answer to Problem 4CLT

  $\begin{array}{l}x=1\\ y=2\\ z=0\end{array}$

Explanation of Solution

Given information:

  $\begin{array}{l}x+y-5z=3\\ 4x+y-11z=6\\ -x-2z=-1\end{array}$

  $x+y-5z=3$ -----(1)

  $4x+y-11z=6$ ---------(2)

  $\begin{array}{l}-x-2z=-1\\ \Rightarrow x=1-2z\end{array}$

  $x=1-2z$ ---------(3)

Subtracting equation-(1) with equation-(2)

We get

  $\begin{array}{l}\Rightarrow x+y-5z-\left(4x+y-11z\right)=3-6\\ \Rightarrow x+y-5z-4x-y+11z=-3\\ \Rightarrow -3x+6z=-3\\ \Rightarrow -x+2z=-1\end{array}$

Substituting value of $x$ from equation-(3)

We get

  $\begin{array}{l}-x+2z=-1\\ \Rightarrow -\left(1-2z\right)+2z=-1\\ \Rightarrow -1+2z+2z=-1\\ \Rightarrow 4z=1-1\\ \Rightarrow 4z=0\\ \Rightarrow z=0\end{array}$

And

Substitute $z=0$ in equation-(3)

We get

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

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

We get

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

Therefore values of $x$ , $y$ and $z$ are

  $\begin{array}{l}x=1\\ y=2\\ z=0\end{array}$