Explanation Procedure used: “ Solving linear systems by Substitution: (1) Solve either of the equations for one variable in terms of the other. (If one of the equations is already in this form, you can skip this step.) (2) Substitute the expression found in step 1 into the other equation. (3) Solve the equation containing one variable. (4) Sometimes variables have been eliminated, and either true statement or false statement results then that means system has infinite solution for true statement or no solution exist for false statement. (5) If neither the case follow the next steps otherwise not. (6) Back-substitute the value found in step 3 into one of the original equations. Simplify and find the value of the remaining variable. (7) Check the proposed solution in both of the system’s given equations”. Calculation: Consider the given equations. − 3 x + 2 y = − 3 ( 1 ) − x + y = 11 6 ( 2 ) Step 1: Isolate y in term of x in the equation (2). − x + y = 11 6 y = 11 6 + x Step 2: Substitute y = 11 6 + x in the equation (1), − 3 x + 2 y = − 3 − 3 x + 2 ( 11 6 + x ) = − 3 Here, the variable y is eliminated. Step 3: Solve the equation obtained in step 2. − 3 x + 2 ( 11 6 + x ) = − 3 − 3 x + 11 3 +2 x = − 3 ( Combine like terms ) − x = − 3 − 11 3 x = 20 3 Step 4: Now, substitute x = 20 3 in one of the given equation and obtain the value of y . Replace x = 20 3 , in the equation (2). − x + y = 11 6 − ( 20 3 ) + y = 11 6 y = 11 6 + 20 3 y = 17 2 Thus, the required solution is ( 20 3 , 17 2 )

4th Edition

ISBN: 9780134556079

Author: Sullivan, Michael, III, Struve, Katherine R., Mazzarella, Janet.

Publisher: Pearson,

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Chapter 4.2, Problem 5E

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To solve: The system of linear equations −3x+2y=−3 and −x+y=116 by substitution.

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Chapter 4.2, Problem 4E

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The system of linear equations − 3 x + 2 y = − 3 and − x + y = 11 6 by substitution.

Solution Summary: The author explains how to solve the system of linear equations by substitution.

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To solve: The system of linear equations −3x+2y=−3 and −x+y=116 by substitution.

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