Stitz-Zeager_College_Algebra_e-book

# 234 117 c 09 009 0009 0 0 0 9 n 1 zeros solution 1

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Unformatted text preview: one of the equations for one variable and substituting it into the other, is full of unpleasantness. Returning to elimination, we note that it is possible to eliminate the troublesome xy term, and the constant term as well, by elimination and doing so we get a more tractable relationship between x and y (E 1) x2 + 2xy − 16 = 0 (E 2) y 2 + 2xy − 16 = 0 Replace E 2 with −− − − −→ −−−−− −E 1 + E 2 (E 1) x2 + 2xy − 16 = 0 (E 2) y 2 − x2 = 0 8.7 Systems of Non-Linear Equations and Inequalities 535 We get y 2 − x2 = 0 or √ = ±x. Substituting y = x into E 1 we get x2 + 2x2 − 16 = 0 so y that x2 = 16 or x = ± 4 3 3 . On the other hand, when we substitute y = −x into E 1, we get 3 √ x2 − 2x2 − 16 = 0 or x2 = −16 which gives no real solutions. Substituting each of x = ± 4 3 3 √ √ √ √ 4343 into the substitution equation y = x yields the solution , − 4 3 3 , − 4 3 3 . We 3,3 leave it to the reader to show that both points satisfy both equatio...
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