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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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 · 2015-05-26T21:23:30

Given
$[1] 2 x + y = 1$ $"[1] "2x+y = 1$ and
$[2] 4 x + 2 y = - 1$ $"[2] "4x+2y = -1$

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

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).

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

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