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anonymous

  • 4 years ago

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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  1. anonymous
    • 4 years ago
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  2. anonymous
    • 4 years ago
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    ok now that i got that out of my system, lets see if we can do it.

  3. Akshay_Budhkar
    • 4 years ago
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    LOL @satellite's attachement

  4. anonymous
    • 4 years ago
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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.

  5. anonymous
    • 4 years ago
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    K thanks

  6. anonymous
    • 4 years ago
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    I think I got it wrong... I think the second equation should be 3/4 (J + 4) = (S + 4)

  7. anonymous
    • 4 years ago
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    looks good to me. i was trying something that i thought might be simpler but it was more complicated

  8. anonymous
    • 4 years ago
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    i will write it anyway, and see if it makes sense

  9. anonymous
    • 4 years ago
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    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

  10. anonymous
    • 4 years ago
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    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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