# How do you solve the system 2x + y = 1 and 4x + 2y = −1 by substitution?

May 26, 2015

Given
$\text{[1] } 2 x + y = 1$ and
$\text{[2] } 4 x + 2 y = - 1$

We would normally rearrange [1] to get an equation for $y$
$\text{[3] } y = 1 - 2 x$
and then substitute this value into equation [2] to solve for $x$

but
Notice something unusual in this case
The left-side of [2] is exactly 2 times the left-side of [1]
but the right-side of [2] is not 2 times the right-side of [1].

The given equations are a pair of parallel lines with no solution (since they do not intersect).