Chapter 7.3, Problem 62E

Solution

To state: The solution for the system of equation using any method:

  $\begin{array}{c}2x+3y+3z+7w=0\\ x+2y+2z+5w=0\\ x+y+2z+3w=-1\end{array}$

The required solution is $\left(w,1-2w,-1-w,w\right)$ .

Given information:

The given system is:

  $\begin{array}{c}2x+3y+3z+7w=0\\ x+2y+2z+5w=0\\ x+y+2z+3w=-1\end{array}$

Explanation:

Consider the given system:

  $\begin{array}{c}2x+3y+3z+7w=0\cdots \left(1\right)\\ x+2y+2z+5w=0\cdots \left(2\right)\\ x+y+2z+3w=-1\cdots \left(3\right)\end{array}$

Multiply equation (3) by -1,

  $\begin{array}{c}\left(x+y+2z+3w=-1\right)\times \left(-1\right)\\ =-x-y-2z-3w=1\cdots \left(4\right)\end{array}$

Now add equations (2) and (4),

  $\begin{array}{l}\underset{¯}{\begin{array}{c}-x-y-2z-3w=1\\ x+2y+2z+5w=0\end{array}}\\ y+2w=1\\ y=1-2w\end{array}$

Multiply equation (2) by -1,

  $\left(x+2y+2z+5w=0\right)\times \left(-2\right)=-2x-4y-4z-10w=0\cdots \left(5\right)$

Now add equations (1) and (5),

  $\begin{array}{l}\underset{¯}{\begin{array}{c}-2x-4y-4z-10w=0\\ 2x+3y+3z+7w=0\end{array}}\\ -y-z-3w=0\end{array}$

Substitute the value of y as $1-2w$ in $-y-z-3w=0$ .

  $\begin{array}{c}-\left(1-2w\right)-z-3w=0\\ -1+2w-z-3w=0\\ -1-w-z=0\\ -w-z=1\\ -z=1+w\\ z=-1-w\end{array}$

Substitute the value of y, and z as $\left(1-2w\right),\text{and}\left(-1-w\right)$ in equation (3),

  $\begin{array}{c}x+y+2z+3w=-1\\ x+\left(1-2w\right)+2\left(-1-w\right)+3w=-1\\ x+1-2w-2-2w+3w=-1\\ x-w=-1+2-1\\ x=w\end{array}$

Therefore, the resultant solution for the system of equation is $\left(w,1-2w,-1-w,w\right)$ .