Question

Solve for${\mathbf{x}}_{\mathbf{1}}$and${\mathbf{x}}_{\mathbf{2}}$in the system of equations given by

$$\begin{gathered} {\mathbf{x}}_{\mathbf{2}}\mathbf{-}\mathbf{3}{\mathbf{x}}_{\mathbf{1}}\mathbf{+}\mathbf{1}\mathbf{.}\mathbf{9}\mathbf{=}\mathbf{0} \\ {\mathbf{x}}_{\mathbf{2}}\mathbf{-}{\mathbf{x}}_{\mathbf{1}}^{\mathbf{2}}\mathbf{+}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{=}\mathbf{0} \end{gathered}$$

using the Gauss method with an initial guess of ${\mathbf{x}}_{\mathbf{1}}\mathbf{=}\mathbf{1}$and${\mathbf{x}}_{\mathbf{1}}\mathbf{=}\mathbf{1}$.

Short answer

The solution of the system are $x_1=1.17$and$x_2=1.62$.

Step-by-step solution

write the given data from question.

The system equations are

$$\begin{gathered} {\mathrm{x}}_2-3{\mathrm{x}}_1+1.9=0 \\ {\mathrm{x}}_2-{\mathrm{x}}_1^2+-3=0 \end{gathered}$$

Consider the given initial condition.

$$\begin{gathered} x_1\left(0\right)=1 \\ {x}_2\left(0\right)=1 \end{gathered}$$

Calculate the solution of the system.

Rewrite the system equation as,

For$x_1$and$x_2$.

$$\begin{gathered} x_1=\frac{1}{3}{x}_2+0.633 \\ {x}_2=3-x_1^2 \end{gathered}$$

By using the initial condition as,

$$\begin{gathered} x_1=\frac{1}{3}{x}_2\left(0\right)+0.633.........\left(1\right) \\ {x}_2\left(1\right)=3-x_1^2\left(0\right)..........\left(2\right) \end{gathered}$$

Substitute 1 for$x_1\left(1\right)$into equation (1).

$$\begin{gathered} x_2\left(1\right)=\frac{1}{3}\left(1\right)+0.633 \\ {x}_2\left(1\right)=0.9667 \end{gathered}$$

Substitute 1 for$x_1\left(0\right)$into equation (2).

$$\begin{gathered} x_2\left(1\right)=3-{\left(1\right)}^2 \\ {x}_2\left(1\right)=2 \end{gathered}$$

Repeat the above process until the 3 digits precision and make table for the results

of iteration.

$$\begin{gathered} x_1\left(n+1\right)=\frac{1}{3}{x}_2\left(n\right)+0.633 \\ {x}_2\left(n+1\right)=3-{\left[x_1\left(n\right)\right]}^2 \end{gathered}$$

After the${50}^{th}$iteration,$x_1$and$x_2$have the precision of 3 digits.

Hence, the solution of the system are $x_1=1.17$and $x_2=1.62$.