We first write the equation

$x = \dfrac{1}{4}[40 + 2y + z]$ ………… (1)

$y = \dfrac{1}{6}[28 + x + 2z]$ ………… (2)

$z = \dfrac{1}{12}[-86 - x + 2y]$ ………… (3)

1.     FIRST ITERATION: We start the approximation y = 0, z = 0, and then we get form (1)

$x_{1} = \dfrac{1}{4}(40) = 10$

  We use this approximation to find y

  Hence we put x = 10, z = 0 in (2 )

$y_{1} = \dfrac{1}{6}[28 + 10 + 2(0)] = 6.3333$

We use these values of$ x_{1}$ and$ y_{1} $to find $z_{1}$

  Hence we put x = 4, y = 6.3333 in (3)

$z_{1} = \dfrac{1}{12}[-86 - 10 + 2(6.3333)] = -6.944$

2.     SECOND ITERATION: We use the latest values of  y and z to find x

Hence we put y = 6.3333, z = -6.9444in (1)

$x_{2} = \dfrac{1}{4}[40 + 2(6.3333) – 6.9444] = 11.4306$

We put x = 11.4306, z = -6.9444 in (2)

$y_{2} = \dfrac{1}{6}[28 + 11.4306 – 2(6.9444)] = 4.2569$

We put x = 11.4306, y = 4.2569 in (3)

$z_{2} = \dfrac{1}{12}[-86 – 11.4306 + 2(4.2569)] = -7.4097$

3.     THIRD  ITERATION: We use the latest values of  y and z to find x

Hence we put y = 4.2569, z = -7.4097 in (1)

$x_{3} = \dfrac{1}{4}[40 + 2(4.2569) – 7.4097] = 10.2760$

We put x = 10.2760, z = -7.4097 in (2)

$y_{3} = \dfrac{1}{6}[28 + 10.2760 – 2(7.4097)] = 3.9094$

We put x = 10.2760, y = 3.9094 in (3)

$z_{3} = \dfrac{1}{12}[-86 – 10.2760 + 2(3.9094)] = -7.3714$

4.     FOURTH  ITERATION: We use the latest values of  y and z to find x

Hence we put y = 3.9094, z = -7.3714 in (1)

$x_{4} = \dfrac{1}{4}[40 + 2(3.9094) – 7.3714] = 10.1118$

We put x = 10.1118, z = -7.3714 in (2)

$y_{4} = \dfrac{1}{6}[28 + 10.1118 – 2(7.3714)] = 3.8948$

We put x = 10.1118, y = 3.8448 in (3)

$z_{4} = \dfrac{1}{12}[-86 – 10.1118 + 2(3.8948)] = -7.3602$

Hence upto two places of decimals x = 10.11, y = 3.89, z = -7.36