Chapter 7, Problem 39RE

To determine

To solve: the system of linear equations and check solutions algebraically.

Answer to Problem 39RE

The solution of the linear equations is $x=1,y=1,z=1,w=0$

Explanation of Solution

Given information:

The system of linear equations is $\{\begin{array}{l}x+4w=1\\ 3y+z-w=4\\ 2y-3w=2\\ 4x-y+2z=5\end{array}$

Concept used:

These are linear equations which are solved by using substitution method.

Calculation:

  $\begin{array}{l}x+4w=1\\ 3y+z-w=4\\ 2y-3w=2\\ 4x-y+2z=5\end{array}$

Let us first isolate x for

  $\begin{array}{l}x+4w=1\\ \Rightarrow x=1-4w\end{array}$

Substitute $x=1-4w$ in the other three equations.

  $\begin{array}{l}2y-3w=2\\ 3y+z-w=4\\ 4\left(1-4w\right)-y+2z=5\end{array}\}\cdots \left(i\right)$

Isolate y for

  $\begin{array}{l}2y-3w=2\\ \Rightarrow y=\frac{2+3w}{2}\end{array}$

Substitute $y=\frac{2+3w}{2}$ in the equation $\left(i\right)$ .

  $\begin{array}{l}3\left(\frac{2+3w}{2}\right)+z-w=4\Rightarrow z+\frac{6+7w}{2}=4\\ 4\left(1-4w\right)-\frac{2+3w}{2}+2z=5\Rightarrow 2z+\frac{-2-35w}{2}+4=5\end{array}\}\cdots \left(ii\right)$

Isolate z for

  $\begin{array}{l}z+\frac{6+7w}{2}=4\\ \Rightarrow z=4-\frac{6+7w}{2}\end{array}$

Substitute this in the equation.

  $\begin{array}{l}2\left(4-\frac{6+7w}{2}\right)+\frac{-2-35w}{2}+4=5\\ \frac{-49w-2}{2}+6=5\end{array}$

On solving

  $\begin{array}{l}\frac{-49w-2}{2}+6=5\\ \Rightarrow -49w-2+12=10\\ \Rightarrow -49w=0\\ \Rightarrow w=0\end{array}$

Substitute $w=0$ to get x,y and z values.

  $\begin{array}{l}z=4-\frac{6+7\left(0\right)}{2}=1\\ y=\frac{2+3\left(0\right)}{2}=1\\ x=1-4\left(0\right)=1\end{array}$

Hence the solution of the linear equations is $x=1,y=1,z=1,w=0$

Check the solution by putting all the values in the given equations

For $x+4w$

  $1+4\left(0\right)=1=\text{R}\text{.H}\text{.S}$ Condition satisfies

For $3y+z-w=4$

  $3\left(1\right)+\left(1\right)-\left(0\right)=4=\text{R}\text{.H}\text{.S}$ Condition satisfies

For $2y-3w=2$

  $2\left(1\right)-3\left(0\right)=2=\text{R}\text{.H}\text{.S}$ Condition satisfies

For $4x-y+2z=5$

  $4\left(1\right)-\left(1\right)+2\left(0\right)=5=\text{R}\text{.H}\text{.S}$ Condition satisfies

Hence $\left(x,y,z,w\right)=\left(1,1,1,0\right)$

Conclusion:

The solution of the linear equations is $x=1,y=1,z=1,w=0$