Chapter 7.3, Problem 40E

To determine

To calculate:Solve the system of linear equations and check any solutions algebraically.

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

Answer to Problem 40E

The solutions for the given linear equations are $x=1,y=0,z=3,w=2$

Explanation of Solution

Given information:

The given pair of linear equations are $\{\begin{array}{c}x+y+z+w=6\\ 2x+3y-w=0\\ -3x+4y+z+2w=4\\ x+2y-z+w=0\end{array}$

Calculation:

These equations are of 4 variables. Let us isolate x from equation 2

From equation 2, we get,

  $x=\frac{-3y-w}{2}$

Substitute x in all the other 3 equations to get,

  $\left[\begin{array}{c}-\frac{3y-w}{2}+y+z+w=6\\ -3\left(-\frac{3y-w}{2}\right)+4y+z+2w=4\\ -\frac{3y-w}{2}+2y-z+w=0\end{array}\right]$

Simplifying it we get,

  $\left[\begin{array}{c}z+\frac{-y+3w}{2}=6\\ z+\frac{17y+w}{2}=4\\ -z+\frac{y+3w}{2}=0\end{array}\right]$

Isolate z from the above first equation we get,

  $z+\frac{-y+3w}{2}=6:z=6+\frac{y+3w}{2}$

Substitute this z value in the other two equations above.

  $\begin{array}{l}\left[\begin{array}{c}6+\frac{y+3w}{2}+\frac{17y+w}{2}=4\\ -\left(6+\frac{y+3w}{2}\right)+\frac{y+3w}{2}=0\end{array}\right]\\ =\left[\begin{array}{c}12+y+3w+17y+w=8\\ -12-y-3w+y+3w=0\end{array}\right]\\ =\left[\begin{array}{c}6+9y-w=4\\ -6+3w=0\end{array}\right]\end{array}$

From this we can get value for w,

  $\begin{array}{l}-6+3w=0\\ 3w=6\\ w=2\end{array}$

Substitute this value to get y,

  $\begin{array}{l}6+9y-w=4\\ 6+9y-2=4\\ 9y=0\\ y=0\end{array}$

Substitute y and w value to get z value,

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

Substitute y,w to get x value.

  $\begin{array}{l}x=\frac{-3y-w}{2}\\ x=-\frac{3\left(0\right)-2}{2}\\ x=1\end{array}$

Conclusion:

The solutions for the given linear equations are $x=1,y=0,z=3,w=2$