$5x-y=9 $ $x=\dfrac {1}{5} (y+9)$ $-x+5y-z=4 $ $y=\dfrac {1}{5} (x+z+4)$ $-y+5z=-6$ $z=\dfrac {1}{5} (y-6)$ First iteration: Put y=z=0, $x_1=\dfrac {9}{5}, $ $y_1 = \dfrac {1}{5}\left ( \dfrac {9}{5}+ 0+4 \right ) = \dfrac {29}{25}, $ $ z_1 = \dfrac {1}{5} \left ( \dfrac {29}{25}-6 \right )= -\dfrac {121}{125}$ Second iteration: $x_2 =\dfrac {1}{5}\left ( \dfrac {29}{25}+9 \right )= \dfrac {254}{125}= 2.032 \ y_2 = \dfrac {1}{5}\left ( \dfrac {254}{125}- \dfrac {121}{125}+ 4 \right ) = \dfrac {633}{625}=1.0128 \ z_2 = \dfrac {1}{5}\left ( \dfrac {633}{625}-6 \right )= -0.99744 $ Third iteration: $x_3= \dfrac {1}{5}(1.0128+9)= 2.00256 \ y_3 = \dfrac {1}{5}(2.00256-0.99744+4)= 1.001024 \ z_3=\dfrac {1}{5} (1.001024-6)= 0.9997952$

Hence the approximate values of (x,y,z) are (2.00256, 1.001024, -0.9997952) which are closed enough to the actual values (2,1,-1).