How do you solve the following linear system 2x-y=4, x+3y=1? ?

Jun 22, 2018

Solve for one term with respect to another, then solve for the second term, then back-solve for the first. You will find that $x = \frac{13}{7}$ and $y = - \frac{2}{7}$

Explanation:

First, we'll want to solve for either $x$ in terms of $y$ or vice versa. For the sake of argument, we'll sove for $x$ in terms of $y$ using the second expression:

$x + 3 y = 1$

$\Rightarrow \textcolor{b l u e}{x = 1 - 3 y}$

Now, we take that solution for $x$ and substitute it in, in expression 1:

$2 \textcolor{b l u e}{x} - y = 4$

$2 \textcolor{b l u e}{\left(1 - 3 y\right)} - y = 4$

$2 - 6 y - y = 4$

$2 - 7 y = 4$

$- 7 y = 2$

$\textcolor{g r e e n}{y = - \frac{2}{7}}$

We now have a solution for $y$. We can then plug it into either expression to solve for $x$:

$x + 3 \textcolor{g r e e n}{y} = 1$

$x + 3 \textcolor{g r e e n}{\left(- \frac{2}{7}\right)} = 1$

$x - \frac{6}{7} = 1$

$x = 1 + \frac{6}{7}$

$\textcolor{g r e e n}{x = \frac{13}{7}}$

And now we have our solution:

$x = \frac{13}{7}$, $y = - \frac{2}{7}$