Question

a. Find all solutions${\mathit{x}}_{\mathbf{1}}\mathbf{,}{\mathit{x}}_{\mathbf{2}}\mathbf{,}{\mathit{x}}_{\mathbf{3}}\mathbf{,}{\mathit{x}}_{\mathbf{4}}$ of the system .

${\mathit{x}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{(}{\mathbf{x}}_{\mathbf{1}}\mathbf{+}{\mathbf{x}}_{\mathbf{3}}\mathbf{)}\mathbf{,}{\mathit{x}}_{\mathbf{3}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{(}{\mathbf{x}}_{\mathbf{2}}\mathbf{+}{\mathbf{x}}_{\mathbf{4}}\mathbf{)}$

b. In partrole="math" localid="1659677484607" $\mathbf{\left(}\mathit{a}\mathbf{\right)}$ , is there a solution with ${\mathit{x}}_{\mathbf{1}}\mathbf{=}\mathbf{1}$and${\mathit{x}}_{\mathbf{4}}\mathbf{=}\mathbf{13}$ ?

Short answer

a. For arbitrary $x_3$and $x_4$, $x_1=3x_3-2x_4,x_2=2x_3-x_4$

b. Yes, $x_1=1,x_2=5,x_3=9,x_4=13$

Step-by-step solution

(a)Step 1: Consider the equations to find other equations

Find the expression for,$x_1$.

$\begin{array}{c}{x}_2=\frac{1}{2}(x_1+x_3)\\ 2x_2=x_1+x_3\\ x_1=2x_2-x_3\mathrm{......}\left(1\right)\end{array}$

Find the expression for, $x_2$.

$\begin{array}{c}{x}_3=\frac{1}{2}(x_2+x_4)\\ 2x_3=x_2+x_4\\ x_2=2x_3-x_4\mathrm{......}\left(2\right)\end{array}$

Express the variable in terms of other variables

Re-write $x_1$in terms of $x_3$and$x_4$ .

$\begin{array}{c}{x}_1=2x_2-x_3\\ x_1=2(2x_3-x_4)-x_3\\ =3x_3-2x_4\mathrm{......}\left(3\right)\end{array}$

Therefore, $x_1=3x_3-2x_4,x_2=2x_3-x_4$

(b)Step 3: Using the above equations, find the values of the variables

Consider the given values of the variables.

$x_1=1$and$x_4=13$

Substitute these values in the above equations to find the values of the remaining variables.

Consider the equation (3).

$\begin{array}{c}{x}_1=3x_3-2x_4\\ 1=3x_3-2\times 13\\ x_3=\frac{27}{3}\\ =9\end{array}$

Consider the equation (2).

$\begin{array}{c}{x}_2=2x_3-x_4\\ =2\times 9-13\\ =18-13\\ =5\end{array}$

Therefore, $x_1=1,x_2=5,x_3=9,x_4=13$.