How do you solve the system?: $x - 4 y = 1, 8 y = 2 x - 2$ $x - 4y =1, 8y = 2x - 2$

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How do you solve the system?: $x - 4 y = 1, 8 y = 2 x - 2$ $x - 4y =1, 8y = 2x - 2$

1 Answer

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Answer 1 · 2015-10-24T11:41:30

They do not have a solution because they are the same straight line

Explanation:

Re-arrange each equation to make y the subject
x - 4y = 1
becomes 4y = x - 1
and y = $\frac{x - 1}{4}$ $(x - 1)/4$ (i)

8y = 2x - 2
which becomes
y = $\frac{2 x - 2}{8}$ $(2x-2)/8$ divide the RHS by 2
to get
y = $\frac{x - 1}{4}$ $(x -1)/4$ (ii)
then you can see that equation (i) = equation (ii)

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How do you solve the system?: $x - 4 y = 1, 8 y = 2 x - 2$ $x - 4y =1, 8y = 2x - 2$

1 Answer

Explanation:

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Science

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How do you solve the system?: x - 4y =1, 8y = 2x - 2x−4y=1,8y=2x−2x - 4y =1, 8y = 2x - 2

1 Answer

Explanation:

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How do you solve the system?: $x - 4 y = 1, 8 y = 2 x - 2$ $x - 4y =1, 8y = 2x - 2$