Chapter 7.5, Problem 78E

To determine

Whether the two system of linear equations yield the same solution, if so find the solution.

Answer to Problem 78E

Both the system of equations has the same solution $x=-1,y=1,z=-3$ .

Explanation of Solution

Given information:$$

  $\text{a)}\{\begin{array}{c}x-2y+z=-6\\ y-5z=16\\ z=-3\end{array}\text{b)}\{\begin{array}{c}x+y-2z=6\\ y+3z=-8\\ z=-3\end{array}$

Solving the given system of equations using Back-substitute method,

Finding solution for $\{\begin{array}{c}x-3y+4z=-11\\ y-z=-4\\ z=2\end{array}$

Substitute $z=2$ in $y-z=-4$ ,

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

Substitute $y=-2$ and $z=2$ in $x-3y+4z=-11$ ,

  $\begin{array}{c}x-3y+4z=-11\\ x-3\left(-2\right)+4\left(2\right)=-11\\ x+14=-11\\ x=-25\end{array}$

The solution for the system of equations $\{\begin{array}{c}x-3y+4z=-11\\ y-z=-4\\ z=2\end{array}$ is $x=-25,y=-2,z=2$ .

Solving the given system of equations using Back-substitute method,

Finding solution for $\{\begin{array}{c}x+4y=-11\\ y+3z=4\\ z=2\end{array}$

Substitute $z=2$ in $y+3z=4$ ,

  $\begin{array}{c}y+3z=4\\ y+3\left(2\right)=4\\ y=-2\end{array}$

Substitute $y=-2$ in $x+4y=-11$ ,

  $\begin{array}{c}x+4y=-11\\ x+4\left(-2\right)=-11\\ x-8=-11\\ x=-3\end{array}$

The solution for the system of equations $\{\begin{array}{c}x+4y=-11\\ y+3z=4\\ z=2\end{array}$ is $x=-25,y=-2,z=2$ .

Thus we can find the given system of equations does not have same solution.