# line meeting curve.

• May 10th 2006, 02:12 AM
line meeting curve.
hi agian!

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

kind regards

• May 10th 2006, 03:47 AM
topsquark
Quote:

hi agian!

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

kind regards

Solve $\displaystyle 2x+3y=1$ for y, plug this y value into $\displaystyle x(x-y) = 2$, 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
• May 10th 2006, 03:50 AM
CaptainBlack
Quote:

hi agian!

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

kind regards

Rewrite the first equation as

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

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

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

and solve for $\displaystyle x$.

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

RonL
• May 10th 2006, 04:00 AM
thanks captainblack! :)
• May 10th 2006, 05:39 AM
re:

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

:(
• May 10th 2006, 05:48 AM
CaptainBlack
Quote:

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

:(

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

which simplifies to:

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

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

RonL