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Math Help - solve simultaneously

  1. #1
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    solve simultaneously

    if someone could help i have an answer for the first one does it look okey. but am unsure how to tackle the 2nd one. many thanks

    (a)

    Log (x) + Log (y) = 2
    y - x = 15

    answer?
    (x=10/y)
    (y = 10/ x)


    (b)
    e^x+y = e^2x+4
    x-y = 4
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  2. #2
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    Quote Originally Posted by decoy808 View Post
    if someone could help i have an answer for the first one does it look okey. but am unsure how to tackle the 2nd one. many thanks

    (a)

    Log (x) + Log (y) = 2
    y - x = 15

    answer?
    (x=10/y)
    (y = 10/ x)


    (b)
    e^x+y = e^2x+4
    x-y = 4
    a) Simultaneous equations should not be solved in terms of each other

    log(xy) = 2

    xy = 100

    x = \frac{100}{y}

    y - \frac{100}{y} = 15 \: \: \longrightarrow \: \: y^2-15y-100=0

    Solve the quadratic for y and then put any valid solutions back into the first equation to check for x

    Spoiler:
    (y-20)(y+5) = 0 \: \: ,\: \therefore \: \: y = 20 \text{or} y=-5

    y = -5 is not in the domain so discard it.

    <br />
log(x) = 2-log(20)

    <br />
x = 10^{2-log(20)} = 5

    Therefore x = 5, y = 20.
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  3. #3
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    Hello, decoy808!

    If I read it correctly, the second one has no solution . . .


    (b)\;\;\begin{array}{cccc}e^{x+y} &=& e^{2x+4} & [1] \\<br />
x-y &=& 4& [2] \end{array}

    Equation [1] gives us: . x+y \:=\:2x+4 \quad\Rightarrow\quad x - y \:=\:-4


    With [2], we have an inconsistent system of equations: . \begin{array}{ccc}x - y &=& 4 \\ x - y &=& \text{-}4\end{array}

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