Explanation Given information: x + 3 y = − 2 4 x 2 + 3 y 2 = 7 Calculation: We have x + 3 y = − 2 4 x 2 + 3 y 2 = 7 From first equation, we get x = − 2 − 3 y Substitute x = − 2 − 3 y in 4 x 2 + 3 y 2 = 7 . 4 ( − 3 y − 2 ) 2 + 3 y 2 = 7 4 ( − 3 y − 2 ) ( − 3 y − 2 ) + 3 y 2 = 7 4 ( 9 y 2 + 6 y + 6 y + 4 ) + 3 y 2 = 7 36 y 2 + 48 y + 16 + 3 y 2 − 7 = 0 39 y 2 + 48 y + 9 = 0 Solving using quadratic formula, we g

Precalculus: A Unit Circle Approach, Books a la Carte Edition plus MyMathLab with Pearson eText -- Access Card Package (2nd Edition)

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Chapter 8.3, Problem 93E

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Find all the points of intersection of the given curves by solving systems of non-linear equations. Sketch the graphs of the curves and show the points of intersection.

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Chapter 8.3, Problem 92E

Chapter 8.3, Problem 94E

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Precalculus: A Unit Circle Approach, Books a la Carte Edition plus MyMathLab with Pearson eText -- Access Card Package (2nd Edition)

Ch. 8.2 - Prob. 1E Ch. 8.2 - Prob. 2E Ch. 8.2 - Prob. 3E Ch. 8.2 - Prob. 4E Ch. 8.2 - Prob. 5E Ch. 8.2 - Prob. 6E Ch. 8.2 - Prob. 7E Ch. 8.2 - Prob. 8E Ch. 8.2 - Prob. 9E Ch. 8.2 - Prob. 10E

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Find all the points of intersection of the given curves by solving systems of non-linear equations. Sketch the graphs of the curves and show the points of intersection.

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Find all the points of intersection of the given curves by solving systems of non-linear equations. Sketch the graphs of the curves and show the points of intersection.

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