Chapter 10.2, Problem 33E

To determine

To calculate:The solution of linear equation

Answer to Problem 33E

  $x=2-2t$ , $y=-\frac{2}{3}+\frac{4}{3}t$ , $z=t$

The equation is linear

Explanation of Solution

Given information:

  $\begin{array}{l}x+3y-2z=0\\ 2x+4z=4\\ 4x+6y=4\end{array}$

Calculation:

Naming the given equations as,

  $\begin{array}{l}x+3y-2z=\mathrm{0........}\left(1\right)\\ 2x+4z=\mathrm{4........}\left(2\right)\\ 4x+6y=\mathrm{4........}\left(3\right)\end{array}$

To eliminate $x$ from equation $\left(2\right)$ ,

Multiplying equation $\left(1\right)$ by $-2$ ,

  $\begin{array}{l}\left(x+3y-2z=0\right)\times \left(-2\right)\\ \left(x+3y-2z\right)\times \left(-2\right)=0\times \left(-2\right)\\ \left(x\times \left(-2\right)\right)+\left(3y\times \left(-2\right)\right)-\left(2z\times \left(-2\right)\right)=0\\ -2x+\left(-6y\right)-\left(-4z\right)=0\\ -2x-6y+4z=0\end{array}$

Naming it as equation $\left(4\right)$ ,

  $-2x-6y+4z=\mathrm{0........}\left(4\right)$

Adding equation $\left(2\right)$ and equation $\left(4\right)$ ,

  $\begin{array}{l}\left(2x+4z=4\right)+\left(-2x-6y+4z=0\right)\\ \left(2x+4z\right)+\left(-2x-6y+4z\right)=4+0\\ \left(2x-2x\right)+\left(-6y\right)+\left(4z+4z\right)=4\\ 0-6y+8z=4\\ -6y+8z=4\end{array}$

Naming it as equation $\left(5\right)$ ,

  $-6y+8z=\mathrm{4........}\left(5\right)$

To eliminate $x$ from equation $\left(3\right)$ ,

Multiplying equation $\left(1\right)$ by $-4$ ,

  $\begin{array}{l}\left(x+3y-2z=0\right)\times \left(-4\right)\\ \left(x\times \left(-4\right)\right)+\left(3y\times \left(-4\right)\right)-\left(2z\times \left(-4\right)\right)=0\times \left(-4\right)\\ -4x+\left(-12y\right)-\left(-8z\right)=0\\ -4x-12y+8z=0\end{array}$

Naming it as equation $\left(6\right)$ ,

  $-4x-12y+8z=\mathrm{0........}\left(6\right)$

Adding equation $\left(3\right)$ and equation $\left(6\right)$ ,

  $\begin{array}{l}\left(4x+6y=4\right)+\left(-4x-12y+8z=0\right)\\ \left(4x+6y\right)+\left(-4x-12y+8z\right)=4+0\\ \left(4x-4x\right)+\left(6y-12y\right)+\left(8z\right)=4\\ 0+\left(-6y\right)+8z=4\\ -6y+8z=4\end{array}$

Naming it as equation $\left(7\right)$ ,

  $-6y+8z=\mathrm{4........}\left(7\right)$

To eliminate $y$ from equation $\left(3\right)$ ,

Multiplying equation $\left(7\right)$ by $-1$ ,

  $\begin{array}{l}\left(-6y+8z=4\right)\times \left(-1\right)\\ \left(-6y\times \left(-1\right)\right)+\left(8z\times \left(-1\right)\right)=4\times \left(-1\right)\\ 6y+\left(-8z\right)=-4\end{array}$

Naming it as equation $\left(8\right)$ ,

  $6y-8z=-\mathrm{4........}\left(8\right)$

Adding equation $\left(7\right)$ and equation $\left(8\right)$ ,

  $\begin{array}{l}\left(-6y+8z=4\right)+\left(6y-8z=-4\right)\\ \left(-6y+8z\right)+\left(6y-8z\right)=4+\left(-4\right)\\ \left(-6y+6y\right)+\left(8z-8z\right)=0\\ 0+0=0\end{array}$

The above equation is true as the values match.

To solve the complete solution of the system,

Solving for $y$ in terms of $z$ ,

Using the equation $\left(7\right)$ ,

  $-6y+8z=4$

Rearranging the above equation to get $y$ in terms of $z$ ,

  $-6y=4-8z$

  $\begin{array}{l}y=-\left(\frac{4-8z}{6}\right)\\ y=-\frac{4}{6}+\frac{8z}{6}\\ y=-\frac{2}{3}+\frac{4}{3}z\end{array}$

Naming it as equation $\left(9\right)$ ,

  $y=-\frac{2}{3}+\frac{4}{3}z\mathrm{........}\left(9\right)$

Solving for $x$ in terms of $z$ ,

Using equation $\left(1\right)$ ,

  $x+3y-2z=0$

Substituting the value in above equation,

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

Naming it as equation $\left(10\right)$ ,

  $x=2-2z\mathrm{........}\left(10\right)$

Assuming that $z=t$ .

Where,

  $t$ is a real number.

Substituting the value $z=t$ in equation $\left(9\right)$ ,

  $y=-\frac{2}{3}+\frac{4}{3}z$

  $y=-\frac{2}{3}+\frac{4}{3}t$

Substituting the value $z=t$ in equation $\left(10\right)$ ,

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

Therefore, the final solution is $\left(2-2t,-\frac{2}{3}+\frac{4}{3}t,t\right)$