line meeting curve.

**dadon** · May 10th 2006, 02:12 AM

hi agian!

how do you fine the coorrdinates where the line $2x+3y=1$ meets the curve $x(x-y) = 2$

kind regards

dadon

**topsquark** · May 10th 2006, 03:47 AM

Originally Posted by dadon

hi agian!

how do you fine the coorrdinates where the line $2x+3y=1$ meets the curve $x(x-y) = 2$

kind regards

dadon

Solve $2x+3y=1$ for y, plug this y value into $x(x-y) = 2$ , and solve the resulting equation for x. The use these x values in either of the original equations to get the corresponding y values. I got (6/5, -7/15) and (-1, 1).

-Dan

**CaptainBlack** · May 10th 2006, 03:50 AM

Originally Posted by dadon

hi agian!

how do you fine the coorrdinates where the line $2x+3y=1$ meets the curve $x(x-y) = 2$

kind regards

dadon

Rewrite the first equation as

$y=\frac{1-2x}{3}$

Then substitute this for $y$ in the second equation:

$x(x-\frac{1-2x}{3}) = 2$

and solve for $x$ .

Then substitute the value/s of $x$ that you have
found back into one of the equations and solve for $y$ .

RonL

**dadon** · May 10th 2006, 04:00 AM

thanks captainblack!

**dadon** · May 10th 2006, 05:39 AM

sorry captainblack about this but i cant seem to solve for x for the following:

$ x(x-\frac{1-2x}{3}) = 2 $

**CaptainBlack** · May 10th 2006, 05:48 AM

Originally Posted by dadon

sorry captainblack about this but i cant seem to solve for x for the following:

$ x(x-\frac{1-2x}{3}) = 2 $

$ x(x-\frac{1-2x}{3})=\frac{x(3x-1+2x)}{3}=\frac{5x^2-x}{3} = 2 $

which simplifies to:

$ 5x^2-x - 6=0 $

which is a quadratic which you can solve using the quadratic formula.

RonL

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