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

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Answer 1 · 2015-10-24T11:41:30

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

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

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

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