Chapter EP, Problem 3.5.9EP

To determine

The solution of the system of equations.

Answer to Problem 3.5.9EP

  $(4,1,2)$ .

Explanation of Solution

Given information:

Solve:

$\begin{array}{l}2x+y-z=7\\ 3x-y+2z=15\\ x-4y+z=2\end{array}$

Calculation:

The given equations are:

  $\begin{array}{l}2x+y-z=7\\ 3x-y+2z=15\\ x-4y+z=2\end{array}$

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

Firstly apply elimination method on equation first and third:

  $\begin{array}{l}\underset{\_}{\begin{array}{l}2x+y-z=7\\ x-4y+z=2\end{array}}\\ 3x-3y=\mathrm{9..............}(1)\end{array}$

Now apply elimination method on equation first and second:

Multiply the first equation by two:

  $\begin{array}{l}\underset{\_}{\begin{array}{l}4x+2y-2z=14\\ 3x-y+2z=15\end{array}}\\ 7x+y=\mathrm{29...................}(2)\end{array}$

Solve the equation $(1)$ and equation $(2)$ by using elimination method:

Multiply equation $(1)$ by seven and $(2)$ by three:

  $\begin{array}{l}\underset{\_}{\begin{array}{l}21x-21y=63\\ 21x+3y=87\\ ---\end{array}}\\ y=1\end{array}$

Now plug the value $y=1$ in given equation $(1)$ :

  $\begin{array}{l}3x-3y=9\\ 3x-3(1)=9\\ x=4\end{array}$

Substitute the value $y=1$ and $x=4$ in given first equation:

  $\begin{array}{l}2x+y-z=7\\ 2(4)+1-z=7\\ -z=7-9\\ z=2\end{array}$

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

  $\begin{array}{l}2x+y-z=7\\ 2(4)+(1)-(2)=7\\ 8+1-2=7\\ 7=7\end{array}$

Thus, the solution is $(4,1,2)$ .