Please Help Me. How do you form an augmented matrix from the system of linear equations and solve.This is Algebra 2. I need help solving questions 6.3 & 6.4. I've read the instructions and I am not understanding what their asking me.​

Answer:   6.3  (x, y) = (3, 1)   6.4  (x, y) = (-4, 2)Step-by-step explanation:An augmented matrix in this context is an array of the coefficients in the equation. For example, the equation x -2y = 3 can be represented by the row of numbers 1 -2 3. When writing the matrix by hand, we usually just separate the elements in a row using blank space. Some calculators require a comma or other character between matrix elements. The entire array is enclosed in square brackets.An augmented matrix may have a vertical line separating the square matrix of coefficients from the rightmost column of constants.Many graphing calculators have functions available for entering and solving linear equations in this form.__6.3 The augmented matrix is ...[tex]\left[\begin{array}{cc|c}2&3&9\\1&-4&-1\end{array}\right][/tex]This array can be used to solve the system of equations using "row operations." The idea is that any row can be multiplied by any number, and any row can be added to any other row (multiplied by some number or not). So, the same techniques used to solve a system by elimination can be used here. We can, for example, subtract twice the second row from the first. Then we have ...[tex]\left[\begin{array}{cc|c}2-2(1)&3-2(-4)&9-2(-1)\\1&-4&-1\end{array}\right]\\\\\left[\begin{array}{cc|c}0&11&11\\1&-4&-1\end{array}\right] \quad\text{simplified}\\\\\left[\begin{array}{cc|c}0&1&1\\1&-4&-1\end{array}\right] \quad\text{first row divided by 11}\\\\\left[\begin{array}{cc|c}0&1&1\\1+4(0)&-4+4(1)&-1+4(1)\end{array}\right] \quad\text{4 times row 1 added to row 2}\\\\\left[\begin{array}{cc|c}0&1&1\\1&0&3\end{array}\right] \quad\text{simplified}[/tex]When we turn this last matrix back into equations, we get ...   0x +y = 1   1x + 0y = 3That is, the solution is (x, y) = (3, 1).__6.4 The augmented matrix is ...[tex]\left[\begin{array}{cc|c}4&5&-6\\3&2&-8\end{array}\right][/tex]Let's start the solution of this one by dividing the first row by 4.[tex]\left[\begin{array}{ccc}1&\frac{5}{4}&-\frac{3}{2}\\3&2&-8\end{array}\right] \\\\\left[\begin{array}{ccc}1&\frac{5}{4}&-\frac{3}{2}\\3-3(1)&2-3(\frac{5}{4})&-8-3(-\frac{3}{2})\end{array}\right] \quad\text{row 2 minus 3 times row 1}\\\\\left[\begin{array}{ccc}1&\frac{5}{4}&-\frac{3}{2}\\0&-\frac{7}{4}&-\frac{7}{2}\end{array}\right] \quad\text{simplify}\\\\\left[\begin{array}{ccc}1&\frac{5}{4}&-\frac{3}{2}\\0&1&2\end{array}\right] \quad\text{multiply row 2 by -4/7}[/tex]We can eliminate the y-term in the first row by subtracting 5/4 times the second row. This will put the matrix in the form that shows us the solution directly.[tex]\left[\begin{array}{ccc}1-\frac{5}{4}(0)&\frac{5}{4}-\frac{5}{4}(1)&-\frac{3}{2}-\frac{5}{4}(2)\\0&1&2\end{array}\right] \quad\text{row 1 minus 5/4 times row 2}\\\\\left[\begin{array}{ccc}1&0&-4\\0&1&2\end{array}\right][/tex]This tells us ...   x = -4   y = 2So the solution is (x, y) = (-4, 2).