Solve the system of linear equations using the Gauss-Jordan elimination method. 5x + 3y = 16 −2x + y = −13 (x, y) =

Answer:The solution for this system is [tex]x = 5, y = 3[/tex].Step-by-step explanation:The Gauss-Jordan elimination method is done by transforming the system's augmented matrix into reduced row-echelon form by means of row operations.We have the following system:[tex]5x + 3y = 16[/tex][tex]-2x + y = -13[/tex]This system has the following augmented matrix.[tex]\left[\begin{array}{ccc}5&3&16\\-2&1&-13\end{array}\right][/tex]The first step is dividing the first line by 5. So:[tex]L_{1} = \frac{L_{1}}{5}[/tex]We now have[tex]\left[\begin{array}{ccc}1&\frac{3}{5}&\frac{16}{5}\\-2&1&-13\end{array}\right][/tex]Now i want to reduce the first row, so I do:[tex]L_{2} = L_{2} + 2L_{1}[/tex]So we have[tex]\left[\begin{array}{ccc}1&\frac{3}{5}&\frac{16}{5}\\0&\frac{11}{5}&-\frac{33}{5}\end{array}\right][\tex].Now, the first step to reduce the second row is:[tex]L_{2} = \frac{5L_{2}}{11}[/tex]So we have:[tex]\left[\begin{array}{ccc}1&\frac{3}{5}&\frac{16}{5}\\0&1&-3\end{array}\right][/tex].Now, to reduce the second row, we do:[tex]L_{1} = L_{1} - \frac{3L_{2}}{5}[/tex]And the augmented matrix is:[tex]\left[\begin{array}{ccc}1&0&5\\0&1&-3\end{array}\right][/tex]The solution for this system is [tex]x = 5, y = 3[/tex].