Chapter 10.2, Problem 25E

To determine

To calculate: The solution of linear equation

Answer to Problem 25E

As the equations are dependent on each other, the equations are linear.

Explanation of Solution

Given information:

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

Calculation:

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

Rearranging the equation $\left(1\right)$ ,

  $y=0+2z$

  $y=2z$

Substitute the value of $y=2z$ in equation $\left(2\right)$ & equation $\left(3\right)$ ,

Equation $\left(2\right)$ becomes,

  $\begin{array}{l}\text{2x}+\left(\text{3}\times \text{2}\right)z=\text{2}\\ \text{2x}+\text{6z}=\text{2}\\ \end{array}$

Equation $\left(3\right)$ becomes,

  $\begin{array}{l}-\text{x}-(2\times \text{2})z+\text{z}=-\text{1}\\ -\text{x}-\text{4z}+\text{z}=-\text{1}\\ -\text{x}-\text{3z}=-\text{1}\end{array}$

Rearranging the above term to get in terms of $\text{x}$ ,

  $-\text{x}=-\text{1}+\text{3z}$

Multiplying both sides by $-\text{1}$ ,

  $\begin{array}{l}\text{-1}\times \text{x}=\left(\text{-1}+\text{3z}\right)\times \left(-\text{1}\right)\\ \text{1x}=\left((-\text{1})\times -\text{1}\right)+\left(\left(\text{3z}\right)\times -1\right)\\ \text{x}=\text{1}-\text{3z}\\ \end{array}$

Substitute the value of $\text{x}=\text{1}-\text{3z}$ in equation $\left(3\right)$ ,

  $\begin{array}{l}2(\text{1}-\text{3z})+\text{6z}=\text{2}\\ \text{2}\times \text{1}+\left(\text{2}\times \left(-\text{3z}\right)\right)+\text{6z}=\text{2}\\ \text{2}-\text{6z}+\text{6z}=\text{2}\\ \text{2}+0=\text{2}\\ 2=2\end{array}$

Since solving the above term,

L.H.S matches R.H.S,

That is,

  $2=2$

Hence,

  $\begin{array}{l}\text{y}=\text{2z}\\ \text{x}=\text{1}-\text{3z}\end{array}$

are true