Answer :
The augmented matrix which represents the system of equations is
[tex]\begin{bmatrix}{3} & {5} & {20} \\ {-1} & {1} & {-4} \\ {} & {} & {}\end{bmatrix}[/tex]Now, lets solve the system by reducing this matrix to row echelon form.
By dividing the first row by 3, we have
[tex]\begin{bmatrix}{1} & {\frac{5}{3}} & {\frac{20}{3}} \\ {-1} & {1} & {-4} \\ {} & {} & {}\end{bmatrix}[/tex]Now, by adding the rows, we get
[tex]\begin{bmatrix}{1} & {\frac{5}{3}} & {\frac{20}{3}} \\ {0} & {\frac{8}{3}} & {\frac{8}{3}} \\ {} & {} & {}\end{bmatrix}[/tex]By multiplying the second row by 8/3, we have
[tex]\begin{bmatrix}{1} & {\frac{5}{3}} & {\frac{20}{3}} \\ {0} & {1} & {1} \\ {} & {} & {}\end{bmatrix}[/tex]and by multiplying the second row by -5/3, we get
[tex]\begin{bmatrix}{1} & {\frac{5}{3}} & {\frac{20}{3}} \\ {0} & {-\frac{5}{3}} & {\frac{5}{3}} \\ {} & {} & {}\end{bmatrix}[/tex]and by adding the second row to the first rwo, we obtain
[tex]\begin{bmatrix}{1} & {0} & {\frac{15}{3}} \\ {0} & {1} & {1} \\ {} & {} & {}\end{bmatrix}[/tex]therefore, the final result is
[tex]\begin{bmatrix}{1} & {0} & {5} \\ {0} & {1} & {1} \\ {} & {} & {}\end{bmatrix}[/tex]This means that the solution is x=5 and y=1. Then, the solution is (5,1)
