Four years ago, Jane was twice as old as Sam. Four years on from now, Sam will be 3/4 of Jane's age. How old is Jane now?

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Four years ago, Jane was twice as old as Sam. Four years on from now, Sam will be 3/4 of Jane's age. How old is Jane now?

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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1 Attachment
ok now that i got that out of my system, lets see if we can do it.
LOL @satellite's attachement

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Let J denote Jane's current age, and S Sam's current age (J - 4) = 2(S - 4) (J + 4) = 3/4 (S + 4) Solve the system for J.
K thanks
I think I got it wrong... I think the second equation should be 3/4 (J + 4) = (S + 4)
looks good to me. i was trying something that i thought might be simpler but it was more complicated
i will write it anyway, and see if it makes sense
Yep I was definitely wrong in my first post. It should be J - 4 = 2(s - 4) 3/4 (J + 4) = S + 4 In which case you should get J=12, S=8
as per marine, \[J-4=2(S-4)\] \[J-4=2S-8\] \[J=2S-4\] is one equation. the next one is \[\frac{3}{4}(J+4)=S+4\] \[\frac{3}{4}J+3=S+4\] \[\frac{3}{4}J-1=S\] now we can substitute \[J=\frac{3}{4}J-1\] for S in the first equation and get \[J=2\times \frac{3}{4}(J-1)-4\] \[J=\frac{3}{2}J-6\] \[6=\frac{1}{2}J\] \[J = 12\]

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