Chapter 9.9, Problem 20WE

(a)

To determine

To find: The general expression for all its solutions.

Answer to Problem 20WE

The general expressions for the solutions of the given system is $(-2z-3,2z+3,z)$

Explanation of Solution

Given:

  $x+2y-2z=3$

  $x+3y-4z=6$$(1)$

  $4x+5y-2z=3$

Calculation:

In the system of equations $(1)$ , adding $-1$ the times of the first equation to second equation, and then adding the $-4$ times to the first equation to find the third equation, the resulting system as follows:

  $\begin{array}{l}x+2y-2z=3\\ y-2z=3\\ -3y+6z=-9\end{array}$

Rewriting the above system of equations in triangular form,

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

In the system of equations $(2)$ , adding second equation to the third equation

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

Now finding values of $x$ and $y$ in terms of $z$ ,

From second equation:

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

From first equation:

  $\begin{array}{l}x+2y-2z=3\\ x+2(2z+3)-2z=3[Replacingy=2z+3]\\ x+4z+6-2z=3\\ x+2z=-3\\ x=-2z-3\end{array}$

Therefore for any number $z$ , the triple $(-2z-3,2z+3,z)$ is a solution

It is shown that the system is dependent. That is value of $x$ and $y$ is dependent on the value of $z$

Conclusion:

The general expressions for the solutions of the given system is $(-2z-3,2z+3,z)$

(b)

To determine

To find: The three particular solutions.

Answer to Problem 20WE

The three solutions of the given systems are $(-5,5,1),(-3,3,0)and(-1,1,-1)$

Explanation of Solution

Given:

  $x+2y-2z=3$

  $x+3y-4z=6$$(1)$

  $4x+5y-2z=3$

Calculation:

The solution is $(-2z-3,2z+3,z)$

For $z=1$ ,

  $\begin{array}{l}x=(-2z-3)\\ =-2(1)-3[Replacingz=1]\\ =-5\\ Again,\\ y=2z+3\\ =2(1)+3[Replacingz=1]\\ =5\end{array}$

Therefore, one particular solution is : $(-5,5,1)$

For $z=0$ ,

  $\begin{array}{l}x=(-2z-3)\\ =-3[Replacingz=0]\\ =-3\\ Again,\\ y=2z+3\\ =0+3[Replacingz=0]\\ =3\end{array}$

Therefore, one particular solution is : $(-3,3,0)$

For $z=-1$ ,

  $\begin{array}{l}x=(-2z-3)\\ =-2(-1)-3[Replacingz=-1]\\ =-1\\ Again,\\ y=2z+3\\ =2(-1)+3[Replacingz=-1]\\ =1\end{array}$

Therefore, one particular solution is : $(-1,1,-1)$

Therefore, three particular solutions are: $(-5,5,1),(-3,3,0)and(-1,1,-1)$

Conclusion:

The three particular solutions are: $(-5,5,1),(-3,3,0)and(-1,1,-1)$