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Math Help - Exponential simultaneous equations

  1. #1
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    Red face Exponential simultaneous equations

    Help with these would be nice =)

    Solve the simultaneous equations:
    2^x - 3^y = 7
    2^x-1 + 3^y+1 = 35

    And another question:
    If 2,1 is a solution to the equations: x^2 + xy + ay = b, and 2ax + 3y = b, find the other solution.

    I did:
    4 + 2 + a = b --(1)
    4a + 3 = b -- (2)
    4a + 3 = 6 + a
    a = 0.6
    b = 6.6
    Then I substituted it into the original equations to get:
    x^2 + xy + 0.6y = 6.6 -- (3)
    1.2x + 3y = 6.6 --(4)

    But I can't get anywhere with it...
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by erika View Post
    Solve the simultaneous equations:
    2^x - 3^y = 7
    2^x-1 + 3^y+1 = 35
    I presume the second equation is actually:
    2^{x - 1} + 3^{y + 1} = 35

    Please use parenthesis!!

    We know that 3^y = 2^x - 7

    So...

    2^{x - 1} + 3^{y + 1} = 35

    (1/2)*2^x + 3*3^y = 35

    (1/2)*2^x + 3*[2^x - 7] = 35

    (1/2)*2^x + 3*2^x - 21 = 35


    (7/2)*2^x = 56

    2^x = (2/7)*56 = 16

    2^x = 2^4

    Thus x = 4 and 3^y = 2^4 - 7 = 16 - 7 = 9

    So y = 2

    -Dan
    Last edited by topsquark; May 6th 2007 at 03:08 PM. Reason: Fixed a boo-boo
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by erika View Post
    Help with these would be nice =)

    Solve the simultaneous equations:
    2^x - 3^y = 7
    2^x-1 + 3^y+1 = 35

    And another question:
    If 2,1 is a solution to the equations: x^2 + xy + ay = b, and 2ax + 3y = b, find the other solution.

    I did:
    4 + 2 + a = b --(1)
    4a + 3 = b -- (2)
    4a + 3 = 6 + a
    You are good to this point.

    4a + 3 = 6 + a

    3a = 3

    Thus a = 1 and b = 6 + a = 7

    So:
    x^2 + xy + y = 7
    2x + 3y = 7

    Solve the bottom equation for y:
    y = (-2/3)x + (7/3)

    Thus
    x^2 + x[(-2/3)x + (7/3)] + [(-2/3)x + (7/3)] = 7

    x^2 + (-2/3)x^2 + (7/3)x + (-2/3)x + (7/3) = 7 <-- Multiply both sides by 3

    3x^2 - 2x^2 + 7x - 2x + 7 = 21

    x^2 + 5x - 14 = 0

    (x - 2)(x + 7) = 0

    Thus x = 2 or x = -7

    For x = 2 ==> y = (-2/3)(2) + (7/3) = -4/3 + 7/3 = 1 (as expected)

    For x = -7 ==> y = (-2/3)(-7) + (7/3) = 14/3 + 7/3 = 21/3 = 7

    So your other solution is (-7, 7).

    -Dan
    Last edited by topsquark; May 6th 2007 at 03:11 PM. Reason: 'Nother mistake. Grrrr....
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  4. #4
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    For the first question:
    if x = 4 and y= 9,
    2^x - 3^y
    = 2^4 - 3^9
    = -19667 (if i didn't see it wrongly?)
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  5. #5
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    Hello, Erika!

    Dan did a great job . . . just slipped on the very last step.


    Solve the simultaneous equations:
    . . [1] .2^x - 3^y .= .7
    . . [2] .2^{x-1} + 3^{y+1} .= .35

    Equation [2] is: .2^x + 33^y .= .35

    Multiply by 2: .2^x + 63^y .= 70
    .Subtract [1]: .2^x .- . 3^y . = .7

    and we have: .73^y .= .63 . . . . 3^y = 9 . . . . y = 2

    Substitute into [1]: .2^x - 3 .= .7 . . . . 2^x = 16 . . . . x = 4

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  6. #6
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    Thanks Dan and Soroban.
    I think there may be a problem with the 2nd question too though:

    The question states: 2x + 3y = 7,
    but using the figures from the answer:
    2(-7) + 3(7/3) = -7

    Sorry about this, I really appreciate the help =)
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  7. #7
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    Hello, Erika!

    If (2,1) is a solution to the equations: x + xy + ay .= .b,
    and: 2ax + 3y .= .b, find the other solution.

    I did:
    4 + 2 + a .= .b . . (1)
    . . 4a + 3 .= .b . . (2)

    . . 4a + 3 .= .6 + a

    Then: .a = 0.6 . . . . no

    You have: .3a .= .3 . . . . a = 1 . . . . b = 7

    The equations become: .x + xy + y .= .7 . [1]
    . . . - . . - . . . . . . . . . . . 2x .+ .3y . = .7 . [2]

    Solve [2] for y: .y .= .(7 - 2x)/3

    Substitute into [1]: .x + x(7 - 2x)/3 + (7 - 2x)/3 .= .7

    Multiply by 3: .3x + 7x - 2x + 7 - 2x .= .21

    We have the quadratic: .x + 5x - 14 .= .0

    . . which factors: .(x - 2)(x + 7) .= .0

    . . and has roots: .x .= .2, -7

    . . and we have: .y .= .1, 7


    We were given the solution (2,1).

    . . The other solution is (-7,7).

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