Solving big systems of linear equations by hand

Question

I have a lot of systems of linear equations to solve by hand like this one: $ x_1 = 1+\frac{1}{4}(x_{1}+x_{2}+x_{4}+x_{5})$
$ x_2 = 1+\frac{1}{4}(x_{1}+x_{2}+x_{3}+x_{6})$
$ x_3 = 1+\frac{1}{4}(x_{2}+x_{3}+x_{4}+x_{7})$
$ x_4 = 1+\frac{1}{4}(x_{1}+x_{3}+x_{4})$
$ x_5 = 1+\frac{1}{4}(x_{1}+x_{5}+x_{6})$
$ x_6 = 1+\frac{1}{4}(x_{2}+x_{5}+x_{6}+x_{7})$
$ x_7 = 1+\frac{1}{4}(x_{3}+x_{6}+x_{7})$
What is the fastest way to solve it? By substitution? By addition? I tried to solve it by substition but it is a very slow method. Thank you.

Answer 1

I don't know if the following way (using addition) is the fastest, but I think it is not bad. Since $x_1=x_3=x_6,x_4=x_5=x_7$, there might be a better way.

$ x_1 = 1+\frac{1}{4}(x_{1}+x_{2}+x_{4}+x_{5})\quad\tag1$
$ x_2 = 1+\frac{1}{4}(x_{1}+x_{2}+x_{3}+x_{6})\quad\tag2$
$ x_3 = 1+\frac{1}{4}(x_{2}+x_{3}+x_{4}+x_{7})\quad\tag3$
$ x_4 = 1+\frac{1}{4}(x_{1}+x_{3}+x_{4})\quad\tag4$
$ x_5 = 1+\frac{1}{4}(x_{1}+x_{5}+x_{6})\quad\tag5$
$ x_6 = 1+\frac{1}{4}(x_{2}+x_{5}+x_{6}+x_{7})\quad\tag6$
$ x_7 = 1+\frac{1}{4}(x_{3}+x_{6}+x_{7})\quad\tag7$

From $(1)+(2)+(3)+(4)+(5)+(6)+(7)$, $$x_1+x_2+x_3+x_4+x_5+x_6+x_7=7+\frac 14(4x_1+4x_2+4x_3+4x_6+3x_4+3x_5+3x_7),$$ i.e. $$x_4+x_5+x_7=28\tag 8$$

From $(4)+(5)+(7)$ and $(8)$, $$x_4+x_5+x_7=3+\frac{1}{4}(2x_1+2x_3+2x_6+x_4+x_5+x_7),$$ i.e. $$x_1+x_3+x_6=36\tag9$$

From $(2)(9)$, $$3x_2=4+x_1+x_3+x_6\Rightarrow \color{red}{x_2=\frac{40}{3}}.$$

From $(1)+(2),(8),(9)$, $$x_1+x_2=2+\frac 14(2x_1+2x_2+x_3+x_4+x_5+x_6)\Rightarrow 9x_1+3x_7=136\tag{10}$$

From $(4)+(5),(8),(9)$, $$x_4+x_5=2+\frac 14(2x_1+x_3+x_4+x_5+x_6)\Rightarrow x_1+3x_7=40\tag{11}$$

From $(10),(11)$, $\color{red}{x_1=12},\color{red}{x_7=\frac{28}{3}}$.

From $(3),(4)$, $\color{red}{x_3=12},\color{red}{x_4=\frac{28}{3}}$.

From $(8),(9)$, $\color{red}{x_5=\frac{28}{3}},\color{red}{x_6=12}$.