91Ó°ÊÓ

First, distribute the coefficients within each parenthesis: \n\n- The first equation becomes \(6x + 3y + 5z = -1\). \n- The second equation becomes \(2x - 6y + 8z = -9\). \n- The third equation becomes \(4 + 4x = -3z + 9y\). The third one is then rearranged to \(4x - 9y + 3z = -4\).

After putting all equations in a standard form, the system of equations can be converted into an augmented matrix: \n\n\[\begin{bmatrix} 6 & 3 & 5 & | & -1 \ 2 & -6 & 8 & | & -9 \ 4 & -9 & 3 & | & -4 \end{bmatrix}\]

By multiplying the first row by \(\frac{1}{3}\) and the second row by \(\frac{1}{2}\) we get the new matrix: \n\n\[\begin{bmatrix} 2 & 1 & \frac{5}{3} & | & -\frac{1}{3} \ 1 & -3 & 4 & | & -\frac{9}{2} \ 4 & -9 & 3 & | & -4 \end{bmatrix}\]\n\nThen, swap rows 1 and 2, and subtract the first row from the third row, 4 times, yielding: \n\n\[\begin{bmatrix} 1 & -3 & 4 & | & -\frac{9}{2} \ 2 & 1 & \frac{5}{3} & | & -\frac{1}{3} \ 0 & 3 & -13 & | & 10 \end{bmatrix}\]\n\nSubtracting the second row 2 times from the first gives: \n\n\[\begin{bmatrix} 0 & -5 & \frac{19}{3} & | & -\frac{35}{6} \ 2 & 1 & \frac{5}{3} & | & -\frac{1}{3} \ 0 & 3 & -13 & | & 10 \end{bmatrix}\]\n\nMultiplying first row by \(-\frac{1}{5}\) and adding 3 times the first row in third one gives the final matrix: \n\n\[\begin{bmatrix} 0 & 1 & -\frac{19}{15} & | & \frac{7}{6} \ 2 & 1 & \frac{5}{3} & | & -\frac{1}{3} \ 0 & 0 & -7 & | & 9 \end{bmatrix}\]

From this matrix form, we can revisit equation form: \n\n- The first equation becomes \(y -\frac{19}{15}z =\frac{7}{6}\). \n- The second equation becomes \(2x + y + \frac{5}{3}z =-\frac{1}{3}\). \n- The third equation becomes \(-7z = 9\). Divide by -7, this gives \(z = -\frac{9}{7}\). \n\nSubstitute \(z\) into the first equation to solve for \(y\) and the resultant value with \(z\) into the second equation to solve for \(x\).