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

Mar 18, 2018

Easiest way is by substitution.

The solution is (x,y)=(–7/5, 3/5).

#### Explanation:

Solving a linear system means finding if there's a point $\left(x , y\right)$ that works for both lines. (There may also be no points, or infinitely many.) In other words, we're going to see if these lines cross, are parallel, or are the same line.

Take the equation $x + 4 y = 1$ and isolate $x$.

$\textcolor{w h i t e}{\implies} x + 4 y = 1$
$\implies x \text{ } = 1 - 4 y$

Now we know that if there's a point $\left(x , y\right)$ that works for both lines, it is the point $\left(x , y\right) = \left(1 - 4 y , \text{ } y\right) .$ Meaning, these coordinates should work for both lines. This is why we now swap $x$ out for $1 - 4 y$ in the other equation:

color(white)=> 2("    "x"     ")+3y=–1
=> 2(1-4y) + 3y = –1
=>"  "2" " -" " 8y " "+ 3y = –1
=>"                "-5y=–3
$\implies \text{ } y = \frac{3}{5}$

So now we know our point $\left(x , y\right)$ occurs when $y = \frac{3}{5}$. What is the $x$-value of this point?

We can now use this new $y$-value in the first equation:

$x = 1 - 4 y$
$\textcolor{w h i t e}{x} = 1 - 4 \left(\frac{3}{5}\right)$
$\textcolor{w h i t e}{x} = 1 - \frac{12}{5}$
color(white)x = –7/5

Thus, our lines cross at the point (x,y) = (–7/5, 3/5). (This is our solution.)