How do solve the following linear system?: $x - 2 y = 8, 4 x = - 11 + y$ $x-2y=8 , 4x=-11+y$ ?

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How do solve the following linear system?: $x - 2 y = 8, 4 x = - 11 + y$ $x-2y=8 , 4x=-11+y$ ?

1 Answer

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Answer 1 · 2016-01-03T19:48:39

I found:
$x = - \frac{30}{7}$ $x=-30/7$
$y = - \frac{43}{7}$ $y=-43/7$

Explanation:

We can isolate $x$ $x$ from the first equation and substitute into the second to find $y$ $y$ :
$x = 8 + 2 y$ $x=8+2y$
So:
$4 (8 + 2 y) = - 11 + y$ $4(8+2y)=-11+y$
$32 + 8 y = - 11 + y$ $32+8y=-11+y$
$7 y = - 43$ $7y=-43$
$y = - \frac{43}{7}$ $y=-43/7$
We use this value back into the first equation:
$x = 8 + 2 (- \frac{43}{7}) = \frac{56 - 86}{7} = - \frac{30}{7}$ $x=8+2(-43/7)=(56-86)/7=-30/7$

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How do solve the following linear system?: $x - 2 y = 8, 4 x = - 11 + y$ $x-2y=8 , 4x=-11+y$ ?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

How do solve the following linear system?: x−2y=8,4x=−11+y x-2y=8 , 4x=-11+y ?

1 Answer

Explanation:

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Impact of this question

How do solve the following linear system?: $x - 2 y = 8, 4 x = - 11 + y$ $x-2y=8 , 4x=-11+y$ ?