Chapter 8.1, Problem 66E

To determine

To solve:The system of linear equations using Gaussian elimination method with back-substitution.

Answer to Problem 66E

Thesolution of the given system is $x=-1\text{and}y=3$ .

Explanation of Solution

Given information:

The system of equations is

  $\{\begin{array}{c}2x+6y=16\\ 2x+3y=7\end{array}$

Calculations:

Writing the given system of linear equations in the augmented matrix form, we have

  $\left[\begin{array}{cccc}2& 6& :& 16\\ 2& 3& :& 7\end{array}\right]$

Performing the row operations to bring it in row-echelon form, we get

  $\begin{array}{l}\left[\begin{array}{cccc}1& 3& :& 8\\ 2& 3& :& 7\end{array}\right]\begin{array}{c}\leftarrow \\ \end{array}\begin{array}{c}{R}_1/2\\ \end{array}\\ \left[\begin{array}{cccc}1& 3& :& 8\\ 0& -3& :& -9\end{array}\right]\begin{array}{c}\\ \leftarrow \end{array}\begin{array}{c}\\ R_2-2R_1\end{array}\\ \left[\begin{array}{cccc}1& 3& :& 8\\ 0& 1& :& 3\end{array}\right]\begin{array}{c}\\ \leftarrow \end{array}\begin{array}{c}\\ R_2/(-3)\end{array}\end{array}$

Now, the matrix is in row-echelon form, and writing it in the system form, we have

  $\{\begin{array}{c}x+3y=8\\ y=3\end{array}$

Substituting $y=3$ in the first equation, we get

  $\begin{array}{l}x+3(3)=8\\ \to x+9=8\\ \Rightarrow x=-1\end{array}$

Conclusion:

Therefore, the solution of the given system is $x=-1\text{and}y=3$ .