Chapter 3.5, Problem 4LC

Solution

To state: The solutions for the system of equation.

  $\left\{\begin{array}{c}2x-y+z=-2\\ x+3y-z=10\\ x+2z=-8\end{array}\right.$

The resultant solution is $\left(2,1,-5\right)$ .

Given information:

The given system equations is:

  $\left\{\begin{array}{c}2x-y+z=-2\\ x+3y-z=10\\ x+2z=-8\end{array}\right.$

Explanation:

Write the original system equations:

  $\left\{\begin{array}{c}2x-y+z=-2\\ x+3y-z=10\\ x+2z=-8\end{array}\right.$

Let’s number the equation,

  $\left\{\begin{array}{c}2x-y+z=-2\cdots \left(1\right)\\ x+3y-z=10\cdots \left(2\right)\\ x+2z=-8\cdots \left(3\right)\end{array}\right.$

Multiply equation (1) by 3,

  $\begin{array}{l}\left(2x-y+z=-2\right)\times 3\\ 6x-3y-3z=-6\cdots \left(4\right)\end{array}$

Add equation (2) and (4) to eliminate variable y ,

  $\begin{array}{r}\left\{\underset{¯}{\begin{array}{c}x+3y-z=10\\ 6x-3y+3z=-6\end{array}}\right.\\ 7x+2z=4\end{array}$

Number the resultant equation,

  $7x+2z=4\cdots \left(5\right)$

Subtract equations (3) and (5),

  $\begin{array}{c}\left\{\underset{¯}{\begin{array}{c}7x+2z=4\\ x+2z=-8\end{array}}\right.\\ 6x=12\end{array}$

  $\begin{array}{l}x=\frac{12}{6}\\ x=2\end{array}$

Now find the value of variable z by substituting the values of x in equation (3) and simplify,

  $\begin{array}{c}x+2z=-8\\ 2+2z=-8\\ 2z=-10\\ z=-5\end{array}$

Now find the value of variable y by substituting the values of x and z in equation (1) and simplify,

  $\begin{array}{c}2x-y+z=-2\\ 2\left(2\right)-y+\left(-5\right)=-2\\ 4-y-5=-2\\ -y=-2+1\\ y=1\end{array}$

Therefore, the solution for the given system of equation is $\left(2,1,-5\right)$ .