How do you solve $y = - 3 x + 4$ $y=-3x+4$ and $- 6 x - 2 y = - 8$ $-6x-2y=-8$ using substitution?

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Answer 1 · 2016-06-26T05:22:13

The two equations are the same. They represent the same line.
There are innumerable number of solutions to these system of equations.

Given -

$y = - 3 x + 4$ $y=-3x+4$ ------------------(1)
$- 6 x - 2 y = - 8$ $-6x-2y=-8$ ----------------(2)

Solve the second equation for $y$ $y$

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

The two equations are the same. They represent the same line.
There are innumerable number of solutions to these system of equations.

How do you solve y=−3x+4y=-3x+4 and −6x−2y=−8 -6x-2y=-8 using substitution?