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    I can solve the first bit, (1,3) and (6, 1/2) but the hence is mandatory so how can I start?

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    Truly grateful to one and all for even a hint...
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    (Original post by Gmart)
    I can solve the first bit, (1,3) and (6, 1/2) but the hence is mandatory so how can I start?

    Name:  Selection_040.jpg
Views: 11
Size:  22.9 KB

    Truly grateful to one and all for even a hint...
    First hint : 2^{b+1} = 2\times 2^b so you have as one of the equations

    3^a + 2\times 2^b = 7

    Look at the original set of equations and think about why this hint is useful. Then try to do something similar with the other equation.
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    (Original post by Gmart)
    I can solve the first bit, (1,3) and (6, 1/2) but the hence is mandatory so how can I start?

    Name:  Selection_040.jpg
Views: 11
Size:  22.9 KB

    Truly grateful to one and all for even a hint...
    Compare the two, might help if you write the two a,b eqs. as 3^a+2(2^b)=7 and (3^a)^2+4(2^b)^2=37

    Then compare against x+2y=7 and x^2+4y^2=37

    Clearly, you have the same situation if you let x=... and y=...
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    (Original post by Notnek)
    First hint : 2^{b+1} = 2\times 2^b so you have as one of the equations

    3^a + 2\times 2^b = 7

    Look at the original set of equations and think about why this hint is useful. Then try to do something similar with the other equation.
    (Original post by RDKGames)
    Compare the two, might help if you write the two a,b eqs. as 3^a+2(2^b)=7 and (3^a)^2+4(2^b)^2=37

    Then compare against x+2y=7 and x^2+4y^2=37

    Clearly, you have the same situation if you let x=... and y=...
    Thank you both - I missed the similarity - maybe bed time !! Got the answer of course now
 
 
 
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