Explanation Given information: The given equation system is: 0.6 x − 0.3 y = − 1.5 0.04 x − 0.02 y = − 0.1 Calculation: The given system equations are: 0.6 x − 0.3 y = − 1.5 ......(1) 0.04 x − 0.02 y = − 0.1 ......(2) Multiply by − 0.04 for equation (1): 0.6 x − 0.3 y = − 1.5 − 0.04 ( 0.6 x ) + 0.04 ( 0.3 y ) = − 0.04 × − 1.5 − 0.024 x + 0.012 y = 0.06 ......(3) Multiply by 0.6 for equation (2): 0.04 x − 0.02 y = − 0.1 0.6 ( 0.04 x ) − 0.6 ( 0.02 y ) = 0.6 × − 0.1 0.024 x + 0.012 y = − 0.06 ......(4) Addition of equation (2) and (3): − 0.024 x + 0.012 y + 0.024 x + 0.012 y = 0.06 − 0.06 0.024 y = 0 y = 0 Put value of y = 0 in equation (1): 0

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ISBN: 9780321978929

Author: Martin-Gay

Publisher: PEARSON

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Chapter 11, Problem 26R

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To solve: The following system by addition method- 0.6x−0.3y=−1.50.04x−0.02y=−0.1 .

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Chapter 11, Problem 25R

Chapter 11, Problem 27R

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Chapter 11 Solutions

EBK ALGEBRA FOUNDATIONS

Ch. 11.1 - Prob. 1VRVC Ch. 11.1 - Prob. 2VRVC Ch. 11.1 - Prob. 3VRVC Ch. 11.1 - Prob. 4VRVC Ch. 11.1 - Prob. 5VRVC Ch. 11.1 - Prob. 6VRVC Ch. 11.1 - Prob. 1ES Ch. 11.1 - Prob. 2ES Ch. 11.1 - Prob. 3ES Ch. 11.1 - Prob. 4ES

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The following system by addition method- 0.6 x − 0.3 y = − 1.5 0.04 x − 0.02 y = − 0.1 .

Solution Summary: The author explains how to solve the following equation system by addition method: l0.6x-0.3y=-1.5

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To solve: The following system by addition method- 0.6x−0.3y=−1.50.04x−0.02y=−0.1 .

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Chapter 11 Solutions