x = -3/5, y = 2, z = 3
Explanation:
Write in matrix form. Find the determinant. Then replace the right side with x, y, and z respectively. Find Ax, Ay, and Az
[tex]\begin{gathered} \begin{bmatrix}{1} & {1} & {0} \\ {0} & {3} & {0} \\ {7} & {6} & {-5}\end{bmatrix}=\text{ }\begin{bmatrix}{3} & & \\ {6} & {} & {} \\ {4} & & {}\end{bmatrix} \\ \end{gathered}[/tex][tex]\begin{gathered} |A|\text{ = }\begin{bmatrix}{1} & {1} & {0} \\ {0} & {3} & {0} \\ {7} & {6} & {-5}\end{bmatrix} \\ \text{ }|A|\text{= 1(-15-0) -1(0-0) + 0(0-21)} \\ |A|\text{ = 1(-15) -1(0) + 0(-21) = -15 -0-0} \\ |A|=\text{ }-15 \end{gathered}[/tex][tex]\begin{gathered} Ax\text{ = }\begin{bmatrix}{3} & {1} & {0} \\ {6} & {3} & {0} \\ {4} & {6} & {-5}\end{bmatrix} \\ Ax\text{ =}3(-15\text{ -0) -1(-30-0)+0(}36-12) \\ Ax\text{ =}3(-15)-1(-30)\text{ +(24) = -45+30+2}4 \\ Ax\text{ =}9 \end{gathered}[/tex][tex]\begin{gathered} A_y=\begin{bmatrix}{1} & {3} & {0} \\ {0} & {6} & {0} \\ {7} & {4} & {-5}\end{bmatrix} \\ A_y=\text{ 1(-30-0) -3(0-0) -0(0-42)} \\ A_y=\text{ 1(-30) -3(0) -0(-42) = -30-0-0 } \\ A_y=\text{ -30} \end{gathered}[/tex][tex]\begin{gathered} A_z=\begin{bmatrix}{1} & {1} & {3} \\ {0} & {3} & {6} \\ {7} & {6} & {4}\end{bmatrix} \\ A_z=\text{ 1(12-36) -1(0-42)+3(0-21)} \\ A_z=\text{ 1(-24)-1(-42) +3(-21) = -24+42-63} \\ A_z=\text{ -4}5 \end{gathered}[/tex][tex]\begin{gathered} x\text{ = }\frac{A_x}{|A|}=\text{ }\frac{9}{-15} \\ x=\text{ }\frac{-3}{5} \\ y\text{ = }\frac{A_y}{|A|}=\text{ }\frac{-30}{-15} \\ y\text{ = 2} \\ z\text{ = }\frac{A_z}{|A|}=\text{ }\frac{-45}{-15} \\ z\text{ = 3} \end{gathered}[/tex]