How do you solve the following linear system:  2x + 4y = 7 , 3x + y = 4 ?

Mar 19, 2016

$\left(\frac{9}{10} , \frac{13}{10}\right)$
Refer below for explanation.

Explanation:

First, arrange them like this so you can compare terms:
$2 x + 4 y = 7$
$3 x + y = 4$

When we solve linear systems like this, we always look for ways to cancel one variable. We can see that if we multiply $3 x + y = 4$ by $- 4$, we will get a $- 4 y$ term. And then, if we add the new equation (with $- 4 y$ in it) to the other equation, the $4 y$ and $- 4 y$ will cancel. Watch:
$- 4 \left(3 x + y = 4\right) \to - 12 x - 4 y = - 16$

$- 12 x \cancel{- 4 y} = - 16$
$+ 2 x \cancel{+ 4 y} = 7$
$- - - - -$
$- 10 x = - 9$
$x = \frac{9}{10}$

We can now use this $x$ value to solve for $y$, like so:
$2 x + 4 y = 7$
$2 \left(\frac{9}{10}\right) + 4 y = 7$
$\frac{9}{5} + 4 y = 7$
$4 y = 7 - \frac{9}{5}$
$4 y = \frac{26}{5}$
$y = \frac{13}{10}$

Therefore our solution is $\left(\frac{9}{10} , \frac{13}{10}\right)$. We can confirm this result in several ways. One, we can substitute these values into the original equations:
$2 x + 4 y = 7 \to 2 \left(\frac{9}{10}\right) + 4 \left(\frac{13}{10}\right) = 7 \to 7 = 7$
$3 x + y = 4 \to 3 \left(\frac{9}{10}\right) + \left(\frac{13}{10}\right) = 4 \to 4 = 4$

We can also look at the graph of these two equations, and confirm that they intersect at the point $\left(\frac{9}{10} , \frac{13}{10}\right)$. Note that $\frac{9}{10} = 0.9$ and $\frac{13}{10} = 1.3$.