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    I'm having some trouble working out the following maths question:

    1a) Given that 3to the power of x = 9 to yhe power of y-1, show that x = 2y -2.

    b) solve the simultaneous equations x = 2y-2
    xsquared = ysquared + 7
    Thanks in advance.
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    1a) 3^x = 9^y-1
    so 3^2(y-1) is the same as that as 3^2 is 9.
    = 3^2y-2
    so x = 2y-2

    b) if x = 2y - 2 you can substitute this into the equation.
    so 2y^2 - 8y + 4 = y^2 + 7
    use the quadratic formula.
    after the beginning steps you should get (8+- root 76)/2
    which simplifies to 4+- 3root2.

    I think that's correct. Sorry if it isnt
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    (Original post by Kay94)
    1a) 3^x = 9^y-1
    so 3^2(y-1) is the same as that as 3^2 is 9.
    = 3^2y-2
    so x = 2y-2

    b) if x = 2y - 2 you can substitute this into the equation.
    so 2y^2 - 8y + 4 = y^2 + 7
    use the quadratic formula.
    after the beginning steps you should get (8+- root 76)/2
    which simplifies to 4+- 3root2.

    I think that's correct. Sorry if it isnt
    Thanks for the help on part 1. But when I used the quadratic formula I got 8 +- 10/ 6

    ...........could you just recheck your answer please and if you still think yours is correct could you explain why.
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    (Original post by Shopaholic94)
    I'm having some trouble working out the following maths question:

    1a) Given that 3to the power of x = 9 to yhe power of y-1, show that x = 2y -2.

    b) solve the simultaneous equations x = 2y-2
    xsquared = ysquared + 7
    Thanks in advance.
    3^x = 9^{y-1}
    3^x = (3^2)^{y-1}
    3^x = 3^{2y-2}

    Cancel the 3 = 3

    x = 2y - 2
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    (Original post by Shopaholic94)
    I'm having some trouble working out the following maths question:

    1a) Given that 3to the power of x = 9 to yhe power of y-1, show that x = 2y -2.

    b) solve the simultaneous equations x = 2y-2
    xsquared = ysquared + 7
    Thanks in advance.
    For part 2:

    x = 2y - 2
    x^2 = y^2 + 7

    So:

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

    This does factorise, no need for quadratic formula:

    (3y + 1)(y - 3)

    So, y = -\frac{1}{3} or y = 3

    Work out x from there...
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    (Original post by Dededex)
    For part 2:

    x = 2y - 2
    x^2 = y^2 + 7

    So:

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

    This does factorise, no need for quadratic formula:

    (3y + 1)(y - 3)

    So, y = -\frac{1}{3} or y = 3

    Work out x from there...
    Thank you so much!
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    (Original post by Shopaholic94)
    Thank you so much!
    No problem
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    (Original post by Shopaholic94)
    Thanks for the help on part 1. But when I used the quadratic formula I got 8 +- 10/ 6

    ...........could you just recheck your answer please and if you still think yours is correct could you explain why.
    Oh yes, apologies, really stupid mistake that I made there lol. Yeah it is 8+- 10/6. So working it out gives y = 3 and y = -1/3. And then obv substitute that into the original x = 2y - 2 equation to give the values of x = 4 and x = -8/3.

    Hope I didn't make any more mistakes there.
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    (Original post by Kay94)
    Oh yes, apologies, really stupid mistake that I made there lol. Yeah it is 8+- 10/6. So working it out gives y = 3 and y = -1/3. And then obv substitute that into the original x = 2y - 2 equation to give the values of x = 4 and x = -8/3.

    Hope I didn't make any more mistakes there.
    Its alright. Thank you anyway!
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    (Original post by Dededex)
    ..
    (Original post by Kay94)
    ..
    Please read the Guide to Posting and in particular what it has to say about posting full solutions.
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    (Original post by DFranklin)
    Please read the Guide to Posting and in particular what it has to say about posting full solutions.
    Sorry should've read that :o:
 
 
 
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