## Elementary Linear Algebra 7th Edition

We can construct the first equation like this: put $x_1$ instead of $t$ and $x_2$ instead of $s$ into the third given equality: $$x_3=3+x_2-x_1$$ and this can be rewritten as $$x_1-x_2+x_3=3.$$ To construct the second equation we have to notice that since there are two arbitrary parameters and three variables we can only have one independent equation which means that the second equation needs to be derived from the first. To do this we will for example take $$-x_1+x_2-x_3=-3$$ which is the first equation multiplied by $-1$. Now we have \begin{align*} x_1-x_2+x_3=&3\\ -x_1+x_2-x_3=&-3 \end{align*} Now let's go to the second part of the problem. To show this we will now equate $x_3=t$ and $x_2=s$. From the first equation it follows by substituting these equalities into it that $$x_1-s+t=3$$ and this means that indeed $$x_1=3+s-t.$$