Explanation Given information: The equation A x + B y + C z = 12 , And it’s three solutions ( 1 , 3 4 , 3 ) , ( 4 3 , 1 , 2 ) and ( 2 , 1 , 1 ) .. Formula used: The system of three linear equations by elimination method is solved as follows: Step 1: First all the equations are written in the standard form as A x + B y + C z = D . Step 2: The decimals or fraction terms are removed by calculations. Step 3: Any one variable is selected to eliminate. Any two equations are selected from the three equations, multiplied and added so that the selected variable gets eliminated and the resulted equation has other variables. Step 4: Different pair of equations are used and same variable as in step 3 is eliminated. Step 5: The system of equations resulting from step 3 and 4 is solved. Step 6: The solution from step 5 is substituted into one of the original three equations and is solved for the third variable with proper application of check. Calculation: Consider the equation as below: A x + B y + C z = 12 And the solutions ( 1 , 3 4 , 3 ) , ( 4 3 , 1 , 2 ) , and ( 2 , 1 , 1 ) . Since, these are the solutions, and then they must satisfy the equation. Substitute the values from each solution in the equation and solve as below: From the ordered solution ( 1 , 3 4 , 3 ) the values are x = 1 , y = 3 4 , z = 3 , substitute and solve as below: A x + B y + C z = 12 A ( 1 ) + B ( 3 4 ) + C ( 3 ) = 12 4 A + 3 B + 12 C = 48 From the ordered solution ( 4 3 , 1 , 2 ) the values are x = 4 3 , y = 1 , z = 2 , substitute and solve as below: A x + B y + C z = 12 A ( 4 3 ) + B ( 1 ) + C ( 2 ) = 12 4 A + 3 B + 6 C = 36 From the ordered solution ( 2 , 1 , 1 ) the values are x = 2 , y = 1 , z = 1 , substitute and solve as below: A x + B y + C z = 12 A ( 2 ) + B ( 1 ) + C ( 1 ) = 12 2 A + B + C = 12 The three equations from the three solutions are as below; 4 A + 3 B + 12 C = 48 ………… (1) 4 A + 3 B + 6 C = 36 ………… (2) And, 2 A + B + C = 12 ………… (3) The equations are already in standard form with no decimals and fractions

Elementary and Intermediate Algebra: Concepts and Applications (7th Edition)

7th Edition

ISBN: 9780134462707

Author: Marvin L. Bittinger, David J. Ellenbogen, Barbara L. Johnson

Publisher: PEARSON

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Chapter 8.4, Problem 56ES

To determine

To calculate: The equation when the provided three solutions are (1,34,3),(43,1,2) and (2,1,1)..

Also, Ax+By+Cz=12.

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Chapter 8.4, Problem 55ES

Chapter 8.4, Problem 57ES

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The equation when the provided three solutions are ( 1 , 3 4 , 3 ) , ( 4 3 , 1 , 2 ) and ( 2 , 1 , 1 ) .. Also, A x + B y + C z = 12 .

Solution Summary: The author explains how the system of three linear equations by elimination method is solved.

Concept explainers

Videos

To calculate: The equation when the provided three solutions are (1,34,3),(43,1,2) and (2,1,1)..

Also, Ax+By+Cz=12.

Want to see the full answer?

Chapter 8 Solutions

The equation when the provided three solutions are ( 1 , 3 4 , 3 ) , ( 4 3 , 1 , 2 ) and ( 2 , 1 , 1 ) .. Also, A x + B y + C z = 12 .

Solution Summary: The author explains how the system of three linear equations by elimination method is solved.

Concept explainers

Videos

To calculate: The equation when the provided three solutions are (1,34,3),(43,1,2) and (2,1,1).. Also, Ax+By+Cz=12.

Want to see the full answer?

Chapter 8 Solutions

To calculate: The equation when the provided three solutions are (1,34,3),(43,1,2) and (2,1,1)..

Also, Ax+By+Cz=12.