Unformatted text preview: = 0 2. x2 + y 2 = 4
4x2 − 9y 2 = 36 4. x2 + y 2 = 4
y − x2 = 0 Solution:
1. Since both equations contain x2 and y 2 only, we can eliminate one of the variables as we did
in Section 8.1.
(E 1)
x2 + y 2 = 4
(E 2) 4x2 + 9y 2 = 36 Replace E 2 with −− − − −→
−−−−−
−4E 1 + E 2 (E 1) x2 + y 2 = 4
(E 2)
5y 2 = 20 From 5y 2 = 20, we get y 2 = 4 or y = ±2. To ﬁnd the associated x values, we substitute each
value of y into one of the equations to ﬁnd the resulting value of x. Choosing x2 + y 2 = 4,
we ﬁnd that for both y = −2 and y = 2, we get x = 0. Our solution is thus {(0, 2), (0, −2)}.
To check this algebraically, we need to show that both points satisfy both of the original
equations. We leave it to the reader to verify this. To check our answer graphically, we sketch
both equations and look for their points of intersection. The graph of x2 + y 2 = 4 is a circle
centered at (0, 0) with a radius of 2, whereas the graph of 4x2 + 9y 2 = 36, when written in the
2
2
standard form x + y4 = 1 is easily recognized as an ellipse centered at (0, 0) with a major
9
axis along the...
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 Fall '13
 Wong
 Algebra, Trigonometry, Cartesian Coordinate System, The Land, The Waves, René Descartes, Euclidean geometry

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