Stitz-Zeager_College_Algebra_e-book

State and prove an analogous result for geometric

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Unformatted text preview: o like terms among the two equations, elimination won’t do us any good. We turn to substitution and from the equation y − 2x = 0, we get y = 2x. Substituting this √ into x2 + y 2 = 4 gives x2 + (2x)2 = 4. Solving, we find 5x2 = 4 or x = ± 2 5 5 . Returning to the equation we used for the substitution, y = 2x, we find y = √ √ √ 45 5 √ 25 , so √ √5 − 2 5 5 , − 4 5 5 . We when x = one solution is 2 5 5 , 4 5 5 . Similarly, we find the other solution to be leave it to the reader that both points satisfy both equations, so that our final answer is √ √ √ √ 2545 , − 2 5 5 , − 4 5 5 . The graph of x2 + y 2 = 4 is our circle from before and the 5,5 graph of y − 2x = 0 is a line through the origin with slope 2. Though we cannot verify the numerical values of the points of intersection from our sketch, we do see that we have two solutions: one in Quadrant I and one in Quadrant III as required. 4. While it may be tempting to solve y − x2 = 0 as y = x2 and substitute, we note that this system is set up for elimination.1 (E 1) x2 + y 2 = 4 (E 2) y − x2 = 0 Replace E 2 with −− − − −→ −−−−...
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