$$ \begin{cases} -x_5+x_4=3\\ -x_5+2x_4+x_3=4\\ 3.5x_4+5x_3+2x_3+2x_2+x_1=3.5 \end{cases} $$ I know how to back substitute to solve for systems where you go you 1 equation down and there is an one extra varible, however with this system there are two extra varibles. How do I solve it then?
I am only just learning about matrices so please use simple language
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considering $x_1,x_2$ as parameters, the system become: $$ \begin{cases} -x_5+x_4=3\\ -x_5+2x_4+x_3=4\\ 3.5x_4+5x_3+2x_3=3.5-2x_2-x_1 \end{cases} $$ Subtracting the first equation from the second we find: $$ x_4+x_3=1 \Rightarrow x_3=1-x_4 $$ and substituting in the third equation we find $$ x_4=1+\dfrac{2x_2+x_1}{1.5} $$ this means that we can give any values to $x_1$ and $x_2$ and we always have a value for $x_4$.
Now using $x_4$ we find the values of $x_3$ and $x_5$ from the other equations.
Backward substitution corresponds to $x_n, x_{n-1}, \ldots x_2, x_1$ in turn. First we see that $x_5$ is a free variable. Then from the first equation we have $$x_4 = 3 + x_5$$ From the second equation we have $$x_3 = 4 - 2x_4 + x_5 = -2 - x_5$$ Then we see that $x_2$ is also a free variable. From the last equation we have $$x_1 = 3.5 - 2x_2 - 2x_3 - 5x_4 - 3.5x_5 = -7.5 - 2x_2 - 5.5x_5$$