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Math Help - Algebra equation

  1. #1
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    Algebra equation

    Hi all,

    I need some help with equations such as this one:

    2(4x + 3)/3 times 3(2x + 7)/4 = 72

    From the above equation this is how far I have reached.

    (8x + 6)/3 times (6x + 21)/4 = 72

    48x^2 + 168x + 36x + 126/12 = 72

    48x^2 + 204x + 126 = 72 times 12 = 864

    48x^2 + 204x = 864 - 126 = 738

    48x^2 + x = 738/204 = 3.617647059

    x^2 + x = 3.617647059/48

    From here I am unsure what to do. Reverse BIDMAS says I will have to square root the square on the x last, but I still have a + x to deal with. They are not like terms so I cannot combine the x^2 + x, and I can't get rid of the square first and add the two x varables because it will mess up the eqaution and go against reverse BIDMAS.

    Thanks
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  2. #2
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    Re: Algebra equation

    Quote Originally Posted by Tygra View Post
    Hi all,

    I need some help with equations such as this one:

    2(4x + 3)/3 times 3(2x + 7)/4 = 72

    From the above equation this is how far I have reached.

    (8x + 6)/3 times (6x + 21)/4 = 72

    48x^2 + 168x + 36x + 126/12 = 72

    48x^2 + 204x + 126 = 72 times 12 = 864

    48x^2 + 204x = 864 - 126 = 738

    48x^2 + x = 738/204 = 3.617647059 <-- please! Never do that again ...

    x^2 + x = 3.617647059/48

    From here I am unsure what to do. Reverse BIDMAS says I will have to square root the square on the x last, but I still have a + x to deal with. They are not like terms so I cannot combine the x^2 + x, and I can't get rid of the square first and add the two x varables because it will mess up the eqaution and go against reverse BIDMAS.

    Thanks
    I assume that your equation is actually

    \displaystyle{\frac{2(4x+3)}3 \cdot \frac{3(2x+7)}4=72}

    Cancel the common factors. Your equation becomes:

    \displaystyle{\frac{(4x+3) \cdot (2x+7)}2=72}

    Now multiply both sides by 2 and expend the brackets. You'll get

    \dispalystyle{8x^2 + 34x - 123 = 0}

    This is a quadratic equation. Use the quadratic formula to solve this equation.
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  3. #3
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    Re: Algebra equation

    Quote Originally Posted by Tygra View Post
    Hi all,

    I need some help with equations such as this one:

    2(4x + 3)/3 times 3(2x + 7)/4 = 72

    From the above equation this is how far I have reached.

    (8x + 6)/3 times (6x + 21)/4 = 72

    48x^2 + 168x + 36x + 126/12 = 72

    48x^2 + 204x + 126 = 72 times 12 = 864

    48x^2 + 204x = 864 - 126 = 738

    48x^2 + x = 738/204 = 3.617647059
    This is incorrect. You have divided the right side by 204 but NOT the left side. That would be
    (48x^2+ 204x)/204= (48/204)x^2+ x.

    x^2 + x = 3.617647059/48

    From here I am unsure what to do. Reverse BIDMAS says I will have to square root the square on the x last, but I still have a + x to deal with. They are not like terms so I cannot combine the x^2 + x, and I can't get rid of the square first and add the two x varables because it will mess up the eqaution and go against reverse BIDMAS.

    Thanks
    Follow Math Help Forum on Facebook and Google+

  4. #4
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    Prove It's Avatar
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    Re: Algebra equation

    Quote Originally Posted by Tygra View Post
    Hi all,

    I need some help with equations such as this one:

    2(4x + 3)/3 times 3(2x + 7)/4 = 72

    From the above equation this is how far I have reached.

    (8x + 6)/3 times (6x + 21)/4 = 72

    48x^2 + 168x + 36x + 126/12 = 72

    48x^2 + 204x + 126 = 72 times 12 = 864

    48x^2 + 204x = 864 - 126 = 738

    48x^2 + x = 738/204 = 3.617647059

    x^2 + x = 3.617647059/48

    From here I am unsure what to do. Reverse BIDMAS says I will have to square root the square on the x last, but I still have a + x to deal with. They are not like terms so I cannot combine the x^2 + x, and I can't get rid of the square first and add the two x varables because it will mess up the eqaution and go against reverse BIDMAS.

    Thanks
    Once your mistake is fixed, as was noted in the other posts, you should get \displaystyle \begin{align*} 8x^2 + 34x - 123 = 0 \end{align*}. You can't use "Reverse BIDMAS" at the moment because that extra x term in the middle makes things problematic. It would be nice if you could write the expression as just a square term and a constant. We can do this using a process known as "Completing the Square". First, you need to make sure your coefficient of \displaystyle \begin{align*} x^2 \end{align*} is 1, so divide both sides by 8 first to give \displaystyle \begin{align*} x^2 + \frac{17}{4}x - \frac{123}{8} = 0 \end{align*}.

    Now, we want to get a single square term of the form \displaystyle \begin{align*} (x + n)^2 \end{align*}. Notice that it expands to \displaystyle \begin{align*} x^2 + 2n\,x + n^2 \end{align*}. Notice that the coefficient of x and the constant term are related, you could get the constant term by dividing the coefficient of x by 2, then squaring. Following this process with our quadratic equation, we have

    \displaystyle \begin{align*} x^2 + \frac{17}{4}x &= \frac{123}{8} \\ x^2 + \frac{17}{4}x + \left( \frac{17}{8} \right)^2 &= \frac{123}{8} + \left( \frac{17}{8} \right)^2 \textrm{ (whatever you add to one side, you have to add to the other to keep the equation balanced, and now on the LHS we have a perfect square...)} \\ \left(x + \frac{17}{8}\right)^2 &= \frac{984}{64} + \frac{289}{64} \\ \left(x + \frac{17}{8}\right)^2 &= \frac{1273}{64} \end{align*}

    You can now solve with "Reverse BIDMAS".
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