Chapter EP, Problem 3.5.6EP

To determine

The solution of the system of equations.

Answer to Problem 3.5.6EP

  $(2,-3,6)$ .

Explanation of Solution

Given information:

Solve:

  $\begin{array}{l}2u-3v=13\\ 3v+w=-3\\ 4u-w=2\end{array}$

Calculation:

The given equations are:

  $\begin{array}{l}2u-3v=13\\ 3v+w=-3\\ 4u-w=2\end{array}$

To solve the system of equations by using the elimination method are as shown below:

Firstly apply elimination method on equation first and third:

Multiply first equation by two:

  $\begin{array}{l}\underset{\_}{\begin{array}{l}4u-6v=26\\ 4u-w=2\\ -\text{ + }-\end{array}}\\ -6v+w=\mathrm{24..............}(1)\end{array}$

Solve the equation $(1)$ and the given second equation by using elimination method:

Multiply second equation by two:

  $\begin{array}{l}\underset{\_}{\begin{array}{l}-6v+w=24\\ 6v+2w=-6\end{array}}\\ w=6\end{array}$

Now plug the value $w=-3$ in given equation third:

  $\begin{array}{l}4u-w=2\\ 4u-6=2\\ 4u=2+6\\ u=2\end{array}$

Substitute the value $u=2$ in original equation one to get the value of $v$ :

  $\begin{array}{l}2u-3v=13\\ 2(2)-3v=13\\ 4-3v=13\\ -3v=13-4\\ v=-3\end{array}$

Now verify the equation by plugging all the values in the original equation:

  $\begin{array}{l}2u-3v=13\\ 2(2)-3(-3)=13\\ 4+9=13\\ 13=13\end{array}$

Thus, the solution is $(2,-3,6)$ .