# Thread: line meeting curve.

1. ## line meeting curve.

hi agian!

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

kind regards

dadon

2. 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

3. 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

4. thanks captainblack!

5. ## re:

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

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

6. 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