## Elementary Linear Algebra 7th Edition

To find the required system we will need to make two equations just in $x_1$ and $x_2$. Put $x_1$ instead of $t$ in the second given equation: $$x_2=3x_1-4$$ which can be rewritten as $$3x_1-x_2=4$$ and there is our first equation. To make our second equation express $t$ in terms of $x_2$ from the second equation given in the problem: $$3t=x_2+4\Rightarrow t=\frac{1}{3}x_2+\frac{4}{3}.$$ Now since $x_1 = t$ we have $$x_1 = \frac{1}{3}x_2+\frac{4}{3}$$ and putting all the variables on the left this becomes $$x_1-\frac{1}{3}x_2 = \frac{4}{3}.$$ Now our system reads \begin{align*} 3x_1-x_2=& 4\\ x_1-\frac{1}{3}x_2 =& \frac{4}{3}. \end{align*} Now let us go to the second part of the problem. To show that this is possible we will put $t$ instead of $x_2$ (because we want $x_2 = t$) to the second equation in the system: $$x_1-\frac{1}{3}t = \frac{4}{3}$$ and this gives $$x_1 = \frac{1}{3}t+\frac{4}{3}$$ and this is what we required.