Explanation Procedure used: The following are the steps to solve the linear systems in three variables by eliminating variables. Step 1: Eliminate the variables one by one from the equations. For that, use any two of the provided three equations to eliminate the selected variable, which makes an equation in two variables. Use the next pair of equations and eliminate the same variables, which leads to another equation in the same two variabes. Step 2: Obtain the value one variable by solving the system of two equations with two variables using the method of addition or substitution. Step 3: Substitute the value of the variable found in Step 2 into any of the equations in two variables to obtain the value of the second variable. Step 4: Get the value of the third variable by substituting the two values obtained in Steps 2 and 3 in any one of the three original equations. Step 5: Check the proposed solution by substituting the values (ordered triple) in the original equations. Calculation: Consider the system of linear equations and express it in the standard form as follows: Frame the first equation in the standard form, x + 3 2 − y − 1 2 + z + 2 4 = 3 2 x + 3 2 − y − 1 2 + z + 2 4 − 3 2 = 0 2 x + 6 − 2 y + 2 + z + 2 − 6 4 = 0 2 x − 2 y + z = − 4 Frame the second equation in the standard form, x − 5 2 + y + 1 3 − z 4 = − 25 6 x − 5 2 + y + 1 3 − z 4 + 25 6 = 0 6 x − 30 + 4 y + 4 − 3 z + 50 12 = 0 6 x + 4 y − 3 z = − 24 Frame the third equation in the standard form, x − 3 4 − y + 1 2 + z − 3 2 = − 5 2 x − 3 4 − y + 1 2 + z − 3 2 + 5 2 = 0 x − 3 − 2 y − 2 + 2 z − 6 + 10 4 = 0 x − 2 y + 2 z = 1 Thus, the system of linear equations is, 2 x − 2 y + z = − 4 (1) 6 x + 4 y − 3 z = − 24 (2) x − 2 y + 2 z = 1 (3) Step 1: Reduce the system to two equations in two variables. Eliminate the variable y add ( 1 ) with (2) ( 1 ) × 2 : 4 x − 4 y + 2 z = − 8 ( 2 ) : 6 x + 4 y − 3 z = − 24 _ 10 x − z = − 32 That is, 10 x − z = − 32 . Now, eliminate y from (2) and (3) by multiplying the equation (3) by 2 and then adding with the equation (2). ( 2 ) : 6 x + 4 y − 3 z = − 24 ( 3 ) × 2 : 2 x − 4 y + 4 z = 2 _ 8 x + z = − 22 That is, 8 x + z = − 22 . Hence, the system of two equations in two variables is obtained. 10 x − z = − 32 (4) 8 x + z = − 22 (5) Step 2: Solve the resulting system of two equations in two variables. Add equation (4) and (5). ( 4 ) : 10 x − z = − 32 ( 5 ) : 8 x + z = − 22 _ 18 x = − 54 Apply the multiplication property of equality and simplify it. 18 x 54 = − 54 18 x = − 3 Step 3: Use back-substitution in one of the equations in two variables to find the value of the second variable. Substitute back x = − 3 in equation (4). 10 ( − 3 ) − z = − 32 − 30 − z = − 32 [ Multiply the terms ] − 30 − z + 30 = − 32 + 30 [ Add 30 from both the sides of the equation ] − z = − 2 z = 2 Step 4: Back-substitute the values found for two variables into one of the originalequations to find the value of the third variable

College Algebra Essentials (5th Edition)

5th Edition

ISBN: 9780134469294

Author: Blitzer, Robert F.

Publisher: PEARSON

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Chapter 5.2, Problem 26E

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The solutions of the system of linear equations {x+32−y−12+z+24=32x−52+y+13−z4=−256x−34−y+12+z−32=−52.

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Chapter 5.2, Problem 25E

Chapter 5.2, Problem 27E

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College Algebra Essentials (5th Edition)

Ch. 5.1 - Consider the system: {2x3y=42x+y=4. Determine if...Ch. 5.1 - Prob. 2CP Ch. 5.1 - Prob. 3CP Ch. 5.1 - Prob. 4CP Ch. 5.1 - Solve the system: {5x2y=410x+4y=7.Ch. 5.1 - Prob. 6CP Ch. 5.1 - Prob. 7CP Ch. 5.1 - Prob. 8CP Ch. 5.1 - Prob. 1CVC Ch. 5.1 - Prob. 2CVC

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The solutions of the system of linear equations { x + 3 2 − y − 1 2 + z + 2 4 = 3 2 x − 5 2 + y + 1 3 − z 4 = − 25 6 x − 3 4 − y + 1 2 + z − 3 2 = − 5 2 .

Solution Summary: The author describes the steps to solve the linear systems in three variables by eliminating variables.

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The solutions of the system of linear equations {x+32−y−12+z+24=32x−52+y+13−z4=−256x−34−y+12+z−32=−52.

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