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# coordinates... watch

1. right i have this question and i can only think that there are no points of intersection with the 2 lines as i cannot factorise it nor does b^2-4ac =0+...(i think)
This is the questions...'the line y=2-3x intersects the curve X^2-y=2 at?

Cheers
xxxxxx
2. (Original post by Smile-Like-You-Mean-It)
right i have this question and i can only think that there are no points of intersection with the 2 lines as i cannot factorise it nor does b^2-4ac =0+...(i think)
This is the questions...'the line y=2-3x intersects the curve X^2-y=2 at?

Cheers
xxxxxx
simultaneous equations, rearrange the first one to equal x, the second one to equal y, then substitute the first one into the second (replace x in the 2nd with whatever you arranged the 1st to)
3. (Original post by IBiot Ash '08)
simultaneous equations, rearrange the first one to equal x, the second one to equal y, then substitute the first one into the second (replace x in the 2nd with whatever you arranged the 1st to)
yeh i did this. But i dont think there is an intercept. Has anyone tried to work it out?
xxxx
4. (Original post by Smile-Like-You-Mean-It)
right i have this question and i can only think that there are no points of intersection with the 2 lines as i cannot factorise it nor does b^2-4ac =0+...(i think)
This is the questions...'the line y=2-3x intersects the curve X^2-y=2 at?

Cheers
xxxxxx
y=2-3x
x^2-y=2

So x^2-(2-3x)=2

work with that.
5. "But i dont think there is an intercept."
Well, there is because it says so in the Question:
"y=2-3x intersects the curve"

I don't really see the difficulty, just solve either x or y (doesn't matter which), this will give you an x or y coordinate, then substitue that value back into one of the original equations to find the x or y you didn't get first time round
6. make the second one y=x^2-2

then x^2-2=2-3x

then x^2+3x-4=0

then (x+4)(x-1)
7. make the second one y=x^2-2

then x^2-2=2-3x

then x^2+3x-4=0

then (x+4)(x-1)

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