To determine To calculate: The solution of the system of equation a − 2 b + 2 c = − 10 2 a − 3 b + 2 c = − 13 a + b + 6 c = − 19 by addition method. Explanation Given Information: The provided system of equation is a − 2 b + 2 c = − 10 2 a − 3 b + 2 c = − 13 a + b + 6 c = − 19 . Formula used: Steps for addition method: Step1. Check whether the equations are in standard form a x + b y + c z = d , if not, and then rewrite the equations in standard form. Step2. Multiply the equations by a constant in such a way that when they are added one variable is eliminated. Step3. Now add the respective sides of the equation. Step4. Simplify the expression obtained in step 3 and compute the value of the variables. Step5. Substitute the value obtained in step 4 in any of the original equation and simplify the expression obtained to obtain the value of the other variables. Step6. Rewrite the solution as an ordered triple. Step7. Verify the solution by substituting the value in all the equations. Calculation: Consider the system of equations, a − 2 b + 2 c = − 10 2 a − 3 b + 2 c = − 13 a + b + 6 c = − 19 Multiply the equation a − 2 b + 2 c = − 10 by ( − 2 ) . ( − 2 ) × ( a − 2 b + 2 c = − 10 ) − 2 a + 4 b − 4 c = 20 Observe that when the equations − 2 a + 4 b − 4 c = 20 , 2 a − 3 b + 2 c = − 13 are added, a gets eliminated. − 2 a + 4 b − 4 c = 20 2 a − 3 b + 2 c = − 13 _ 0 a + b − 2 c = 7 _ Hence, the sum of the equations − 2 a + 4 b − 4 c = 20 , 2 a − 3 b + 2 c = − 13 is b − 2 c = 7 . Now multiply the equation a + b + 6 c = − 19 by − 2 . ( − 2 ) × ( a + b + 6 c = − 19 ) − 2 a − 2 b − 12 c = 38 Observe that when the equations 2 a − 3 b + 2 c = − 13 , − 2 a − 2 b − 12 c = 38 are added, a gets eliminated. 2 a − 3 b + 2 c = − 13 − 2 a − 2 b − 12 c = 38 _ 0 a − 5 b − 10 c = 25 _ Hence, the sum of the equations 2 a − 3 b + 2 c = − 13 , − 2 a − 2 b − 12 c = 38 is − 5 b − 10 c = 25 . Now multiply the equation b − 2 c = 7 by 5 . 5 × ( b − 2 c = 7 ) 5 b − 10 c = 35 Observe that when the equations 5 b − 10 c = 35 , − 5 b − 10 c = 25 are added, b gets eliminated. 5 b − 10 c = 35 − 5 b − 10 c = 25 _ 0 b − 20 c = 60 _ Solve the equation − 20 c = 60 for c , c = 60 − 20 c = − 3 Substitute − 3 for c in the equation b − 2 c = 7

Intermediate Algebra For College Students (10th Edition)

10th Edition

ISBN: 9780134758992

Author: Allen R. Angel, Dennis Runde

Publisher: PEARSON

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Chapter 4.2, Problem 22ES

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To calculate: The solution of the system of equation a−2b+2c=−102a−3b+2c=−13a+b+6c=−19 by addition method.

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Chapter 4.2, Problem 21ES

Chapter 4.2, Problem 23ES

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