Question #55c9f

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Question #55c9f

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How do you solve the system of equations $2 x - 5 y = 10$ $2x - 5y = 10$ and $4 x - 10 y = 20$ $4x - 10y = 20$ ?
How do you solve the system of equations $2 x - 3 y = 6$ $2x-3y=6$ and $3 y - 2 x = - 6$ $3y-2x=-6$ ?

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Answer 1 · 2017-10-18T23:23:26

(4,2) or x=4 and y=2

Explanation:

Since you are given the second equation solved in terms of x, you can plug what x equals in for x in the first equation to get rid of the x variable. This would be:
$5 (4 y - 4) - 4 y = 12$ $5color(red)((4y-4))-4y=12$
You would then distribute and simplify:
$20 y - 20 - 4 y = 12$ $20y-20-4y=12$
$16 y - 20 = 12$ $16y-20=12$
$16 y = 32$ $16y=32$
$y = 2$ $y=2$
You then plug this y value back into one of the original equations to find x. Let's use the second equation:
$x = 4 (2) - 4$ $x=4color(red)((2))-4$
So $x = 4$ $x=4$
And the overall solution, in coordinate form, would be: (4,2)

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Question #55c9f

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