How do you solve $x - 2 y = - 3$ $x-2y= -3$ and $x + 6 y = 1$ $x+6y= 1$ using substitution?

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Answer 1 · 2018-05-01T10:39:53

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

You can rearrange either the first or second equation making $x$ $x$ the subject.
Rearranging the first equation:

$x - 2 y = - 3$ $x-2y=-3$
$x = 2 y - 3$ $x=2y-3$

Substitute this $x$ $x$ equation into the $x$ $x$ in the second equation:

$x + 6 y = 1$ $x+6y=1$
$(2 y - 3) + 6 y = 1$ $(2y-3)+6y=1$
$8 y - 3 = 1$ $8y-3=1$
$8 y = 4$ $8y=4$
$y = \frac{1}{2}$ $y=1/2$

Then substitute this $y$ $y$ equation into either the $x$ $x$ equation:

$x = 2 y - 3$ $x=2y-3$
$x = 2 (\frac{1}{2}) - 3$ $x=2(1/2)-3$
$x = 1 - 3$ $x=1-3$
$x = - 2$ $x=-2$

How do you solve x-2y= -3x−2y=−3x-2y= -3 and x+6y= 1x+6y=1x+6y= 1 using substitution?