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Math Help - AS question on powers, sim eq, and inequalities

  1. #1
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    AS question on powers, sim eq, and inequalities

    These are some questions from a AS Core 1 Practise Paper and I don't have a clue what to do.I had a go at B a few times and got a few different answers based around root7.

    A) Given that 3^x = 9^(y-1) , show that x=2y-2.

    B) Solve the simulatious eqs, x=2y-2
    x^2=(y^2)+7

    C) Find a set of values for x for which 6x-7 < x+3 and (2x^2)+7x -4 >0

    I had a go at C and solved each equation, x<2 and -4<x<1/2 but wasn't sure what to do when it says find a set of values for both.

    Sorry I'm not sure how to write powers properly on computors
    Thanks
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  2. #2
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    Quote Originally Posted by markcantdomaths View Post
    A) Given that 3^x = 9^(y-1) , show that x=2y-2.
    Note that 9^{y - 1} = \left( {3^2 } \right)^{y - 1} = 3^{2(y - 1)} .

    Quote Originally Posted by markcantdomaths View Post
    B) Solve the simulatious eqs, x=2y-2
    x^2=(y^2)+7
    Substitute the first equation into the second. You'll get a quadratic equation that you should be able to tackle.

    Quote Originally Posted by markcantdomaths View Post
    C) Find a set of values for x for which 6x-7 < x+3 and (2x^2)+7x -4 >0

    I had a go at C and solved each equation, x<2 and -4<x<1/2 but wasn't sure what to do when it says find a set of values for both.
    Did you do a sketch? That'd be helpful.
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  3. #3
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    Oh that was actually quite simple, just the x and y powers scared me! Thanks.
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  4. #4
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    Quote Originally Posted by markcantdomaths View Post
    These are some questions from a AS Core 1 Practise Paper and I don't have a clue what to do.I had a go at B a few times and got a few different answers based around root7.

    A) Given that 3^x = 9^(y-1) , show that x=2y-2.

    B) Solve the simulatious eqs, x=2y-2
    x^2=(y^2)+7

    ...
    Hello,

    to A):

    3^x = 9^{y-1}~\iff~3^x = \left( \left(3 \right)^2 \right)^{y-1}~\iff~3^x = 3^{2x-2}

    Since the bases are equal the exponents must be equal too. Thus

    x = 2y-2

    to B) Plug in the term for x into the 2nd equation:

    (2y-2)^2 = y^2 + 7~\implies~ 4y^2-8y+4=y^2+7~\implies~ 3y^2-8y-3=0

    Solve this quadratic equation for y. I've got:

    y = 3~\vee~y = -\frac13 Plug in these values into the first equation to calculate x:
    x = 4~\vee~x=-\frac83
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