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

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Answer 1 · 2015-05-31T22:27:39

We have the following system :

$2 x + y = 1$ $2x+y=1$
$4 x + 2 y = - 1$ $4x+2y=-1$

We need to isolate $x$ $x$ or $y$ $y$ on one of the equations. Let's isolate $y$ $y$ on the first one, it's the easiest :

$2 x + y = 1 \Leftrightarrow y = 1 - 2 x$ $2x+y = 1 hArr y = 1-2x$

Now we can replace the $y$ $y$ on the second equation :

$4 x + 2 y = 4 x + 2 (1 - 2 x) = 4 x - 4 x + 2 = 2 \neq - 1$ $4x+2y=4x+2(1-2x)=4x-4x+2=2 != -1$ .

There aren't any solutions to that system.

How do you solve the system 2x + y = 1 2x+y=12x + y = 1 and 4x + 2y = −14x+2y=−14x + 2y = −1 by substitution?