Explanation Given Information: The provided expression is 1 ( x + y ) 2 − 1 x 2 y . Formula used: The difference of two rational expression: A B − C B = A − C B To subtract when the denominators are the same, subtract the second numerators from the first and keep the common denominators. The algebraic identity for differences of squares of two numbers: ( a 2 − b 2 ) = ( a + b ) ( a − b ) Calculation: Consider the expression 1 ( x + y ) 2 − 1 x 2 y . The denominators of the terms 1 ( x + y ) 2 − 1 x 2 are ( x + y ) 2 and x 2 . Therefore, multiply by x 2 x 2 in the first term and ( x + y ) 2 ( x + y ) 2 in the second term. 1 ( x + y ) 2 − 1 x 2 = x 2 x 2 ⋅ 1 ( x + y ) 2 − ( x + y ) 2 ( x + y ) 2 ⋅ 1 x 2 = x 2 x 2 ( x + y ) 2 − ( x + y ) 2 x 2 ( x + y ) 2 Apply the difference of two rational expression 1 ( x + y ) 2 − 1 x 2 = x 2 x 2 ⋅ 1 ( x + y ) 2 − ( x + y ) 2 ( x + y ) 2 ⋅ 1 x 2 = x 2 x 2 ( x + y ) 2 − ( x + y ) 2 x 2 ( x + y ) 2 = x 2 − ( x + y ) 2 x 2 ( x + y ) 2 Now, apply the algebraic identity for differences of squares of two numbers, 1 ( x + y ) 2 − 1 x 2 = x 2 x 2 ⋅ 1 ( x + y ) 2 − (

7th Edition

ISBN: 9781337291262

Author: Stefan Waner, Steven Costenoble

Publisher: Brooks Cole

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Chapter 0.4, Problem 15E

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To calculate: The simplified form of the expression 1(x+y)2−1x2y.

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Chapter 0.4, Problem 14E

Chapter 0.4, Problem 16E

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Chapter 0 Solutions

Finite Mathematics, Loose-leaf Version

Ch. 0.1 - Prob. 1E Ch. 0.1 - Prob. 2E Ch. 0.1 - Prob. 3E Ch. 0.1 - Prob. 4E Ch. 0.1 - Prob. 5E Ch. 0.1 - Prob. 6E Ch. 0.1 - Prob. 7E Ch. 0.1 - Prob. 8E Ch. 0.1 - Prob. 9E Ch. 0.1 - Prob. 10E

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The simplified form of the expression 1 ( x + y ) 2 − 1 x 2 y .

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