Step 1

Given system of equations are

\(5x+3y+z=-8\)...(1)

\(x-3y+2z=20\)...(2)

\(14x-2y+3z=20\)...(3)

Step 2

By subtracting 5 times equation (2) from equation (1), we get

\(5x+3y+z-5(x-3y+2z)=-8-100\)

\(18y-9z=-108\)

\(2y-y=-12\)...(4)

Step 3

By subtracting 14 times equation (2) from equation (3), we get

\(14x-2y+3z-14(x-3+2z)=20-280\)

\(40y-25z=-260\)

\(8y-5z=-52\)...(5)

Step 4

By subtracting 4 times equation (4) from equation (5), we get

\(8y-5z-4(2y-z)=-52+48\)

\(-z=-4\)

\(z=4\)

Step 5

By substituting \(z=4\) in equation (4), we get value of y and then substituting values of y and z in equation (2), we get value of x

\(2y-z=-12\)

\(2y-4=-12\)

\(2y=-8\)

\(y=-4\)

\(x-3y+2z=20\)

\(x-3(-4)+2(4)=20\)

\(x+12+8=20\)

\(x=0\)

Step 6

Answer: Solution is \(x=0\),\(y=-4\) and \(z=4\)