Determine the number of solutions to the following system. xy = 2 x^2 = 3 + y^2
In order to tackle this problem, you want both equations to equal y. Always keep this in mind. So, the first one could be changed to let y be by itself by dividing both sides by x, so it becomes \[y=2/x\] Then for the next one, the 3 has to be subtracted from both sides, and then the square root has to be taken of both sides, resulting in \[y=\sqrt{2x-3}\] Since this proves that both of these equation represent y, they both must be equal to each other. So set them equal to each other! \[\sqrt{2x-3}=2/x\] Bring all the terms together, set it equal to 0, and solve for x! Simple enough. :)
can you walk me thru that last step? how do you bring all the terms together?
When you do AC Method, in a regular quadratic, its like this \[ax^2+ax+a\], So you try to find factors such that multiply to get a but add to get ax. Check out the drawing.|dw:1341473645111:dw| The ones that I underlined, that's the kinda thing you have to do, especially towards the end as well.
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