Explanation Given Information: The provided system of equations are: 4 a 2 − 25 b 2 = 0 And 2a 2 − 10 b 2 = 3 b + 4 Calculation: Consider the linear equations, 4 a 2 − 25 b 2 = 0 …… ( 1 ) And 2a 2 − 10 b 2 = 3 b + 4 …… ( 2 ) Now multiply equation ( 2 ) by − 2 , -4a 2 + 20 b 2 = − 6 b − 8 …… ( 3 ) Now add both the equations ( 1 ) and ( 3 ) , and solve ( 4 a 2 − 4 a 2 ) + ( − 25 b 2 + 20 b 2 ) = − 6 b − 8 − 5 b 2 = − 6 b − 8 5 b 2 − 6 b − 8 = 0 ( 5 b + 4 ) ( b − 2 ) = 0 Further simplify the factors, 5 b + 4 = 0 5 b = − 4 b = − 4 5 And b − 2 = 0 b = 2 Now substitute b = − 4 5 in the equations ( 1 ) and solve for a : 4 a 2 − 25 b 2 = 0 4 a 2 − 25 ( − 4 5 ) 2 = 0 4 a 2 − 16 = 0 a 2 = 16 4 Further simplify, a 2 = 4 a = ± 2 The ordered pair is ( 2 , − 4 5 ) and ( − 2 , − 4 5 ) . Now substitute b = 2 in the equation, ( 1 ) and solve for a : 4 a 2 − 25 b 2 = 0 4 a 2 − 25 ( 2 ) 2 = 0 4 a 2 − 100 = 0 a 2 = 100 4 Further simplify, a 2 = 25 a = ± 5 The ordered pair is ( 5 , 2 ) and ( − 5 , 2 ) . The following graph serves as a check and intersection on this graph occurs at ( 2 , − 4 5 ) , ( − 2 , − 4 5 ) , ( 5 , 2 ) and ( − 5 , 2 ) . Check: For the ordered pair ( 2 , − 4 5 ) : Consider equation ( 1 ) , 4 a 2 − 25 b 2 = 0 4 ( 2 ) 2 − 25 ( − 4 5 ) 2 = ? 0 16 − 16 = ? 0 0 = 0 Now, consider equation ( 2 )

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ISBN: 9780321848741

Author: Marvin L. Bittinger

Publisher: PEARSON

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Chapter 13.4, Problem 43ES

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To calculate: The solution of the equations 4a2−25b2=0 and 2a2−10b2=3b+4.

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Chapter 13.4, Problem 42ES

Chapter 13.4, Problem 44ES

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Chapter 13 Solutions

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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. 1ES Ch. 13.1 - Prob. 2ES Ch. 13.1 - Choose from the following list the word that best...Ch. 13.1 - Choose from the following list the word that best...Ch. 13.1 - Prob. 5ES

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The solution of the equations 4 a 2 − 25 b 2 = 0 and 2a 2 − 10 b 2 = 3 b + 4 .

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To calculate: The solution of the equations 4a2−25b2=0 and 2a2−10b2=3b+4.

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Chapter 13 Solutions