Explanation Given Information: The provided equations are, y − 2 x = − 6 2 y − x = 5 Formula used: Multiplication property of equality: A = B and A ⋅ C = B ⋅ C are equivalent where C ≠ 0 . If A , B and C are real numbers then, Addition property of equality: A = B and A + C = B + C are equivalent. Calculation: Consider the system of equations: y − 2 x = − 6 …… (1) 2 y − x = 5 …… (2) Since, coefficient of y is 1 in the equation ( 1 ) Use addition property, Add ( 2 x ) to both sides of the equation ( 1 ) Thus, y − 2 x + 2 x = − 6 + 2 x y = − 6 + 2 x Put the value y = − 6 + 2 x in equation ( 2 ) Thus, 2 ( − 6 + 2 x ) − x = 5 Remove parentheses, − 12 + 4 x − x = 5 Take like terms together, − 12 + 3 x = 5 Use addition property, Add ( 12 ) to both sides, − 12 + 3 x + 12 = 5 + 12 3 x = 17 Use multiplication property, Multiply both side by ( 1 3 ) . 3 x ( 1 3 ) = 17 ( 1 3 ) x = 17 3 Put x = 17 3 in equation ( 2 ) 2 y − 17 3 = 5 Use addition property, Add ( 17 3 ) to both sides, 2 y − 17 3 + 17 3 = 5 + 17 3 2 y = 32 3 Use multiplication property, Multiply both side by ( 1 2 )

9th Edition

ISBN: 9780321997173

Author: Marvin L. Bittinger, Judith A. Beecher

Publisher: PEARSON

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Chapter 13.2, Problem 15ES

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To calculate: The solution of equations y−2x=−6, 2y−x=5 using substitution method. First solve one equation for one variable..

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Chapter 13.2, Problem 14ES

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Ch. 13.1 - Prob. 1SR Ch. 13.1 - Prob. 2SR Ch. 13.1 - Prob. 1DE Ch. 13.1 - Prob. 2DE Ch. 13.1 - Prob. 3DE Ch. 13.1 - Prob. 4DE Ch. 13.1 - Prob. 5DE Ch. 13.1 - Prob. 6DE Ch. 13.1 - Prob. 7DE Ch. 13.1 - Prob. 8DE

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The solution of equations y − 2 x = − 6 , 2 y − x = 5 using substitution method. First solve one equation for one variable..

Solution Summary: The author calculates the solution of the equations y-2x=-6,2y-x =5 using substitution method.

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To calculate: The solution of equations y−2x=−6, 2y−x=5 using substitution method. First solve one equation for one variable..

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Chapter 13 Solutions