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Thread: 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

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

    Cancel the common factors. Your equation becomes:

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

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

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

    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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    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 $\displaystyle \displaystyle \begin{align*} 8x^2 + 34x - 123 = 0 \end{align*}$. 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 $\displaystyle \displaystyle \begin{align*} x^2 \end{align*}$ is 1, so divide both sides by 8 first to give $\displaystyle \displaystyle \begin{align*} x^2 + \frac{17}{4}x - \frac{123}{8} = 0 \end{align*}$.

    Now, we want to get a single square term of the form $\displaystyle \displaystyle \begin{align*} (x + n)^2 \end{align*}$. Notice that it expands to $\displaystyle \displaystyle \begin{align*} x^2 + 2n\,x + n^2 \end{align*}$. 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

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

    You can now solve with "Reverse BIDMAS".
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