# solving this equation.

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• Sep 8th 2008, 07:50 AM
jvignacio
solving this equation.
$\displaystyle y - 2 = 2 \cdot \sqrt{3}(x - \frac{\pi}{3})$

any ideas, just confused with the gradient
• Sep 8th 2008, 07:56 AM
Krizalid
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?
• Sep 8th 2008, 07:56 AM
Simplicity
Quote:

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

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.
• Sep 8th 2008, 08:04 AM
jvignacio
Quote:

Originally Posted by Air
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?
• Sep 8th 2008, 08:06 AM
jvignacio
Quote:

Originally Posted by Krizalid
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 ;)
• Sep 8th 2008, 08:13 AM
Simplicity
Quote:

Originally Posted by jvignacio
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 $\displaystyle y=mx+c$.
• Sep 9th 2008, 03:41 AM
jvignacio
Quote:

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

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!