Explanation Given Information: The equations are: x 2 + y 2 = 10 And x y = 3 Calculation: Consider the linear equations, x 2 + y 2 = 10 …… ( 1 ) And x y = 3 …… ( 2 ) Now simplify above equation, y = 3 x …… ( 3 ) Now substitute y = 3 x in the equations ( 1 ) and solve for x , x 2 + y 2 = 10 x 2 + ( 3 x ) 2 = 10 x 2 + 9 x 2 = 10 Further, multiply by x 2 on both sides, x 4 + 9 = 10 x 2 x 4 − 10 x 2 + 9 = 0 ( x 2 − 1 ) ( x 2 − 9 ) = 0 Now simplify both the factors, x 2 − 1 = 0 x 2 = 1 x = ± 1 And, x 2 − 9 = 0 x 2 = 9 x = ± 3 Now substitute x = 1 in the equation, ( 3 ) and solve for y : y = 3 x = 3 1 = 3 The ordered pair is ( 1 , 3 ) . Now substitute x = − 1 in the equation, ( 3 ) and solve for y : y = 3 x = 3 − 1 = − 3 The ordered pair is ( − 1 , − 3 ) . Now substitute x = 3 in the equation, ( 3 ) and solve for y : y = 3 x = 3 3 = 1 The ordered pair is ( 3 , 1 ) . Now substitute x = − 3 in the equation, ( 3 ) and solve for y : y = 3 x = 3 − 3 = − 1 The ordered pair is ( − 3 , − 1 ) . The following graph serves as a check and intersection on this graph occurs at ( 1 , 3 ) , ( − 1 , − 3 ) , ( 3 , 1 ) and ( − 3 , − 1 ) . Check: For the ordered pair ( 1 , 3 ) : Consider equation ( 1 ) , x 2 + y 2 = 10 ( 1 ) 2 + ( 3 ) 2 = ? 10 1 + 9 = ? 10 10 = 10 Now, consider equation ( 2 )

Pearson eText Elementary and Intermediate Algebra: Concepts and Applications -- Instant Access (Pearson+)

7th Edition

ISBN: 9780137361847

Author: Marvin Bittinger, David Ellenbogen

Publisher: PEARSON+

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Chapter 13.4, Problem 37ES

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To calculate: The solution of the equations x2+y2=10, xy=3

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Chapter 13.4, Problem 36ES

Chapter 13.4, Problem 38ES

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Pearson eText Elementary and Intermediate Algebra: Concepts and Applications -- Instant Access (Pearson+)

Ch. 13.1 - Prob. 1YT Ch. 13.1 - Prob. 2YT Ch. 13.1 - Prob. 3YT Ch. 13.1 - Prob. 4YT Ch. 13.1 - Prob. 5YT Ch. 13.1 - Prob. 1CYU Ch. 13.1 - Prob. 2CYU Ch. 13.1 - Prob. 3CYU Ch. 13.1 - Prob. 4CYU Ch. 13.1 - Prob. 5CYU

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The solution of the equations x 2 + y 2 = 10 , x y = 3

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Chapter 13 Solutions