How do you solve $3 x + y = 1$ $3x + y = 1$ and $x – y = 3$ using substitution?

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Answer 1 · 2018-05-23T13:40:49

$(x,y) = (1,-2)$

From the second argument, we can come to conclusion that

$x = 3 + y$

Therefore, we have:

$3(3 + y) + y = 1$

Furthermore:

$9 + 3y + y = 1$

$4y = -8$

$y=-2$

Back to the second equation:

$x = 3 + y$

$x = 3 + (-2)$

$x = 1$

So

$(x,y) = (1,-2)$

How do you solve 3x + y = 13x+y=13x + y = 1 and x – y = 3x – y = 3 using substitution?