How do you solve the system of equations $- 10 x + 8 y = 2$ $-10x+8y=2$ and $- 2 x - 4 y = 6$ $-2x-4y=6$ by elimination?

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Answer 1 · 2016-11-01T08:48:35

$x = - 1 = y$ $x=-1=y$

$- 10 x + 8 y = 2$ $-10x+8y=2$ $(i)$ $(i)$

$- 2 x - 4 y = 6$ $-2x-4y=6$ $(i i)$ $(ii)$

So we have 2 equations in 2 unknowns, a system that is in principle solvable.

If I add $(i) + 2 \times (i i)$ $(i)+2xx(ii)$ , we get:

$- 10 x + 8 y - 4 x - 8 y = 2 + 12$ $-10x+8y-4x-8y=2+12$

i.e. $- 14 x = 14$ $-14x=14$ ,

$x = \frac{- 14}{14} = - 1$ $x=(-14)/(14)=-1$

And, $y = - 1$ $y=-1$

How do you solve the system of equations −10x+8y=2-10x+8y=2 and −2x−4y=6-2x-4y=6 by elimination?