Solve the system of equations by finding the reduced row echelon form of the augmented matrix. Write the solutions for x and y in terms of z comma where z can be any number. StartLayout left-brace 1st Row 1st Column 3 x plus y plus 4 z 2nd Column equals 3rd Column -3 2nd Row 1st Column negative x plus y plus 4 z 2nd Column equals 3rd Column 17 EndLayout

Answer:The solutions for the given system of equations are:[tex]\left \{ {{x=-5} \atop {y=12-4z}} \right.[/tex]Step-by-step explanation:Given the equation system:[tex]\left \{ {{3x+y+4z=-3} \atop {-x+y+4z=17}} \right.[/tex]We obtain the following matrix:[tex]\left[\begin{array}{cccc}3&1&4&-3\\-1&1&4&17\end{array}\right][/tex]Step 1: Multiply the fisrt row by 1/3.[tex]\left[\begin{array}{cccc}1&\frac{1}{3} &\frac{4}{3}&-1\\-1&1&4&17\end{array}\right][/tex]Step 2: Sum the first row and the second row.[tex]\left[\begin{array}{cccc}1&\frac{1}{3} &\frac{4}{3}&-1\\0&\frac{4}{3} &\frac{16}{3}&16\end{array}\right][/tex]Step 3: Multiply the second row by 3/4.[tex]\left[\begin{array}{cccc}1&\frac{1}{3} &\frac{4}{3}&-1\\0&1 &4&12\end{array}\right][/tex]Step 4: Multiply the second row by -1/3 and sum the the first row.[tex]\left[\begin{array}{cccc}1&0 &0&-5\\0&1 &4&12\end{array}\right][/tex]The result of the reduced matrix is:[tex]\left \{ {{x=-5} \atop {y+4z=12}} \right.[/tex]This is equal to:[tex]\left \{ {{x=-5} \atop {y=12-4z}} \right.[/tex]These are the solutions for the system of equations in terms of z, where z can be any number.