Chapter 7.3, Problem 61E

Solution

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

  $\begin{array}{c}2x+y+z+4w=-1\\ x+2y+z+w=1\\ x+y+z+2w=0\end{array}$

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

Given information:

The given system is:

  $\begin{array}{c}2x+y+z+4w=-1\\ x+2y+z+w=1\\ x+y+z+2w=0\end{array}$

Explanation:

Consider the given system:

  $\begin{array}{c}2x+y+z+4w=-1\cdots \left(1\right)\\ x+2y+z+w=1\cdots \left(2\right)\\ x+y+z+2w=0\cdots \left(3\right)\end{array}$

Multiply equation (1) by -1,

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

Now add equations (3) and (4),

  $\begin{array}{l}\underset{¯}{\begin{array}{c}-2x-y-z-4w=1\\ x+y+z+2w=0\end{array}}\\ -x-2w=1\\ -x=1+2w\\ x=-1-2w\end{array}$

Multiply equation (3) by -1,

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

Now add equations (2) and (5),

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

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

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

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