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Thread: solving this equation.

  1. #1
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    solving this equation.

    $\displaystyle y - 2 = 2 \cdot \sqrt{3}(x - \frac{\pi}{3})$

    any ideas, just confused with the gradient
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  2. #2
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    Solve for $\displaystyle y$ or $\displaystyle x?$ This is a recent question that belongs to other thread, so I'll assume that you want to isolate $\displaystyle y:$

    $\displaystyle y-2=2\sqrt{3}\left( x-\frac{\pi }{3} \right)\implies y=2\sqrt{3}x-\frac{2\sqrt{3}}{3}\pi +2=2\sqrt{3}x-\frac{2}{\sqrt{3}}\pi +2.$

    Does that make sense?
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  3. #3
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    Quote Originally Posted by jvignacio View Post
    $\displaystyle y - 2 = 2 \cdot \sqrt{3}(x - \frac{\pi}{3})$

    any ideas, just confused with the gradient
    An equation of a line is in the form $\displaystyle y = mx + c$

    Where:
    • $\displaystyle y, x$ = Variables
    • $\displaystyle m$ = Gradient
    • $\displaystyle c$ = $\displaystyle y$-intercept
    $\displaystyle y - 2 = 2\sqrt{3}\left(x-\frac{\pi}{3}\right)$
    $\displaystyle y-2 = 2\sqrt{3}x - \frac{2\pi\sqrt{3}}{3}$
    $\displaystyle y= 2\sqrt{3}x - \frac{2\pi\sqrt{3}}{3} + 2$
    $\displaystyle y = 2\sqrt{3}x - \frac{2\pi\sqrt{3} + 6}{3}$

    ^ Compare this form with $\displaystyle y = mx + c$ to work out what your gradiant $\displaystyle \left(m\right)$ is.
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  4. #4
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    Quote Originally Posted by Air View Post
    An equation of a line is in the form $\displaystyle y = mx + c$


    Where:
    • $\displaystyle y, x$ = Variables
    • $\displaystyle m$ = Gradient
    • $\displaystyle c$ = $\displaystyle y$-intercept
    $\displaystyle y - 2 = 2\sqrt{3}\left(x-\frac{\pi}{3}\right)$
    $\displaystyle y-2 = 2\sqrt{3}x - \frac{2\pi\sqrt{3}}{3}$
    $\displaystyle y= 2\sqrt{3}x - \frac{2\pi\sqrt{3}}{3} + 2$
    $\displaystyle y = 2\sqrt{3}x - \frac{2\pi\sqrt{3} + 6}{3}$

    ^ Compare this form with $\displaystyle y = mx + c$ to work out what your gradiant $\displaystyle \left(m\right)$ is.

    how did u get +6 on the numerator at the end?
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  5. #5
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    Quote Originally Posted by Krizalid View Post
    Solve for $\displaystyle y$ or $\displaystyle x?$ This is a recent question that belongs to other thread, so I'll assume that you want to isolate $\displaystyle y:$

    $\displaystyle y-2=2\sqrt{3}\left( x-\frac{\pi }{3} \right)\implies y=2\sqrt{3}x-\frac{2\sqrt{3}}{3}\pi +2=2\sqrt{3}x-\frac{2}{\sqrt{3}}\pi +2.$

    Does that make sense?
    yeah it was for solving y!
    makes sence now. gracias mate
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  6. #6
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    Quote Originally Posted by jvignacio View Post
    how did u get +6 on the numerator at the end?
    $\displaystyle y = 2\sqrt{3}x - \frac{2\pi\sqrt{3} + 6}{3}$

    Is...

    $\displaystyle y = 2\sqrt{3}x - \frac{2\pi\sqrt{3}}{3} + \frac{6}{3}$

    Which is...

    $\displaystyle y = 2\sqrt{3}x - \frac{2\pi\sqrt{3}}{3} + 2$

    By keeping it as $\displaystyle y = 2\sqrt{3}x - \frac{2\pi\sqrt{3} + 6}{3}$, I though it would be easier for you to compare the form with $\displaystyle y=mx+c$.
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  7. #7
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    Quote Originally Posted by Air View Post
    $\displaystyle y = 2\sqrt{3}x - \frac{2\pi\sqrt{3} + 6}{3}$

    Is...

    $\displaystyle y = 2\sqrt{3}x - \frac{2\pi\sqrt{3}}{3} + \frac{6}{3}$

    Which is...

    $\displaystyle y = 2\sqrt{3}x - \frac{2\pi\sqrt{3}}{3} + 2$

    By keeping it as $\displaystyle y = 2\sqrt{3}x - \frac{2\pi\sqrt{3} + 6}{3}$, I though it would be easier for you to compare the form with $\displaystyle y=mx+c$.
    yes yes now i understand, cheers man!
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