HELP PLEASE!!! for this system, what are the coordinates of the solution that lies in quadrant i? x^2+4y^2=100 and 4y-x^2=-20

Question

HELP PLEASE!!! for this system, what are the coordinates of the solution that lies in quadrant i?  x^2+4y^2=100 and 4y-x^2=-20

anonymous · Answer

$$x ^{2}+4y ^{2}=100$$
$$4y-x ^{2}=-20$$

anonymous · Answer

so you want to make it into a equation where y is on one side and x on the other. After that, you plug those equation into the graphing calculator. Then you can find the solutions in quandrant i which is the top left square.

anonymous · Answer

so for x^2+4y^2=100. You would square root both sides to simplify which gives you x+4y=100. Then you minus 4y from both sides to get x on one side and y on another. so you get x=4y+100. You want y to end up by itself so you minus 100 from both sides and get x-100=4y. Then you divide both sides by 4 to get y by itself. Your equation should be x/4-25=y.

anonymous · Answer

thank you sam!