{-3 x+5 y+6 z=-8 3 x+y+z=-4 x-4 y-2 z=-9.

To solve the system of linear equations, we can use the method of substitution or elimination. Let's use the elimination method to solve this system.<br /><br />First, let's multiply the second equation by 3 to make the coefficients of x in the first and second equations equal:<br /><br />\( \left\{\begin{array}{l}-3 x+5 y+6 z=-8 \\ 9 x+3 y+3 z=-12 \\ x-4 y-2 z=-9\end{array}\right. \)<br /><br />Next, let's add the first and second equations to eliminate x:<br /><br />\( \left\{\begin{array}{l}6 y+9 z=-20 \\ x-4 y-2 z=-9\end{array}\right. \)<br /><br />Now, let's multiply the second equation by 3 to make the coefficients of x in the second and third equations equal:<br /><br />\( \left\{\begin{array}{l}6 y+9 z=-20 \\ 3 x-12 y-6 z=-27\end{array}\right. \)<br /><br />Next, let's add the first and second equations to eliminate x:<br /><br />\( \left\{\begin{array}{l}6 y+9 z=-20 \\ -6 y+3 z=-27\end{array}\right. \)<br /><br />Now, let's add the two equations to eliminate y:<br /><br />\( 12 z=-47 \)<br /><br />Dividing both sides by 12, we get:<br /><br />\( z=-\frac{47}{12} \)<br /><br />Now, let's substitute the value of z into the first equation:<br /><br />\( -3 x+5 y+6(-\frac{47}{12})=-8 \)<br /><br />Simplifying, we get:<br /><br />\( -3 x+5 y-\frac{47}{2}=-8 \)<br /><br />Multiplying both sides by 2 to eliminate the fraction, we get:<br /><br />\( -6 x+10 y-47=-16 \)<br /><br />Now, let's substitute the value of z into the second equation:<br /><br />\( 3 x+y+(-\frac{47}{12})=-4 \)<br /><br />Simplifying, we get:<br /><br />\( 3 x+y-\frac{47}{12}=-4 \)<br /><br />Multiplying both sides by 12 to eliminate the fraction, we get:<br /><br />\( 36 x+12 y-47=-48 \)<br /><br />Now, we have a system of two equations with two variables:<br /><br />\( \left\{\begin{array}{l}-6 x+10 y-47=-16 \\ 36 x+12 y-47=-48\end{array}\right. \)<br /><br />We can solve this system by multiplying the first equation by 6 and the second equation by -1, then adding the two equations to eliminate x:<br /><br />\( \left\{\begin{array}{l}-36 x+60 y-282=-96 \\ -36 x-12 y+47=48\end{array}\right. \)<br /><br />Adding the two equations, we get:<br /><br />\( 48 y-235=-48 \)<br /><br />Dividing both sides by 48, we get:<br /><br />\( y=-\frac{235}{48} \)<br /><br />Now, let's substitute the value of y into the first equation:<br /><br />\( -6 x+10(-\frac{235}{48})-47=-16 \)<br /><br />Simplifying, we get:<br /><br />\( -6 x-\frac{1175}{24}-47=-16 \)<br /><br />Multiplying both sides by 24 to eliminate the fraction, we get:<br /><br />\( -144 x-1175-1128=-384 \)<br /><br />Simplifying, we get:<br /><br />\( -144 x-2303=-384 \)<br /><br />Adding 2303 to both sides, we get:<br /><br />\( -144 x=1919 \)<br /><br />Dividing both sides by -144, we get:<br /><br />\( x=-\frac{1919}{144} \)<br /><br />Therefore, the solution to the system of equations is:<br /><br />\( x=-\frac{1919}{144}, y=-\frac{235}{48}, z=-\frac{47}{12} \)