# How do you solve the system of equations: 2x + y = 1 and 4x + 2y = −1?

Aug 15, 2015

$\left\{\begin{matrix}x = \emptyset \\ y = \emptyset\end{matrix}\right.$

#### Explanation:

Start by writing the system as given to you

$\left\{\begin{matrix}2 x + y = 1 \\ 4 x + 2 y = - 1\end{matrix}\right.$

Notice that you can simplify the second equation by dividing all the terms by $2$ to get

$\frac{4}{2} \cdot x + \frac{2}{2} \cdot y = - \frac{1}{2}$

This is equivalent to

$2 x + y = - \frac{1}{2}$

Notice that the left side of the second equation is identical to the left side of the first equation, but that this expression equals two different values, $1$ and $- \frac{1}{2}$, respectively.

In other words, you have

$\left\{\begin{matrix}2 x + y = 1 \\ 2 x + y = - \frac{1}{2}\end{matrix}\right.$

Let's say that you wanted to solve this system by substitution

$y = 1 - 2 x$

$2 x + \left(1 - 2 x\right) = - \frac{1}{2}$

$\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{2 x}}} + 1 - \textcolor{b l u e}{\cancel{\textcolor{b l a c k}{2 x}}} = - \frac{1}{2}$

$1 \textcolor{red}{\ne} - \frac{1}{2}$

Since $1 \ne - \frac{1}{2}$ for any value of $x$ and of $y$, the system of equations has no real solution. You're essentially dealing with two parralel lines that will never intersect to produce a valid solution.