Explanation Given information: The system of equation is, u − v + 6 w = 8 3 u − v + 6 w = 14 − u − 2 v − 3 w = 7 Formula used: Follow the steps below to reach the solution as, Step-1: Write all the equations in the standard form as A x + B y + C z = D . Step-2: If there are any decimals or fraction terms, remove them by calculations. Step-3: Choose any one variable to eliminate. Select any two equations from the three equations, and multiply and add them so that the selected variable gets eliminated and the resulted equation has other variables. Step-4: Select different pair of equations and repeat step 3, to get one more equation in the same two variables. Step-5: Solve the system of equation in 2 variables of the equation got in step 3 and 4. Step-6: From step 5, value for the two variables are found, find the value for the third variable by substituting the values of other two in any of the equations. Calculation: Consider the system of equations, u − v + 6 w = 8 …… (1) And, 3 u − v + 6 w = 14 …… (2) And, − u − 2 v − 3 w = 7 …… (3) The equations are already in standard form with no decimals and fractions. So, to eliminate one variable, make the coefficient of u opposite, multiply equation (1) by − 3 and add to equation (2) as, − 3 u + 3 v − 18 w = − 24 − 3 u + 3 v − 18 w + 3 u − v + 6 w = − 24 + 14 Combine the like terms and simplify as, − 3 u + 3 v − 18 w + 3 u − v + 6 w = − 24 + 14 2 v − 12 w = − 10 v − 6 w = − 5 Now, add equation (1) and (3), to eliminate u as, u − v + 6 w = 8 u − v + 6 w − u − 2 v − 3 w = 8 + 7 Solve by combining the like terms as, u − v + 6 w − u − 2 v − 3 w = 8 + 7 − 3 v + 3 w = 15 − v + w = 5 Then the new system of equation in two variables is, v − 6 w = − 5 …… (4) And, − v + w = 5 …… (5) Now, solve the above system of equation in two variables as, Add the two equations (4) and (5) as, v − 6 w = − 5 v − 6 w − v + w = − 5 + 5 Solve by combining the like terms as, v − 6 w − v + w = − 5 + 5 − 5 w = 0 w = 0 Substitute the value of w in any of the equations, here equation (4) and solve for the value of b as, v − 6 w = − 5 v − 6 ( 0 ) = − 5 v = − 5 Again, substitute the value of w and v in any of the equations, here equation (1) and solve for value of u as, u − v + 6 w = 8 u − ( − 5 ) + 6 ( 0 ) = 8 u + 5 + 0 = 8 u = 3 Therefore, the values are u = 3 , v = − 5 and w = 0

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 20ES

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To calculate: The solution for the system of equations u−v+6w=8, 3u−v+6w=14 and −u−2v−3w=7.

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Chapter 8.4, Problem 19ES

Chapter 8.4, Problem 21ES

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

Ch. 8.1 - In 2016, there were 35 species of amphibians in...Ch. 8.1 - 2. Refer to Example 2 . For another necklace, Star...Ch. 8.1 - Prob. 3YT Ch. 8.1 - Prob. 4YT Ch. 8.1 - Prob. 1CYU Ch. 8.1 - Prob. 2CYU Ch. 8.1 - Prob. 3CYU Ch. 8.1 - Prob. 4CYU Ch. 8.1 - Prob. 1ES Ch. 8.1 - Prob. 2ES

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The solution for the system of equations u − v + 6 w = 8 , 3 u − v + 6 w = 14 and − u − 2 v − 3 w = 7 .

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To calculate: The solution for the system of equations u−v+6w=8, 3u−v+6w=14 and −u−2v−3w=7.

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