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 will be eliminated, and either true statement or false statement results. Then, the system has infinite solution for true statement or no solution exists for false statement. (5) If neither the case, follow the next steps otherwise not. (6) Substitute back 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 given equations”. Calculation: The given equations are, y = − 5 6 x + 3 (1) y = − 4 3 x (2) Step 1: Since both the equations are already in the form of y in terms of x , no need to isolate the variable. Step 2: Substitute − 4 3 x for y in the equation (1). y = − 5 6 x + 3 ( − 4 3 x ) = − 5 6 x + 3 Here, the variable y is eliminated. Step 3: Solve the equation obtained in step 2 as follows. ( − 4 3 x ) = − 5 6 x + 3 − 4 3 x + 5 6 x = 3 ( Combine like terms ) − 3 6 x = 3 x = − 6 Step 4: Now, substitute − 6 for x in one of the given equation and obtain the value of y . Replace x by − 6 in the equation (2). y = − 4 3 x = − 4 3 ( − 6 ) = 8 Thus, the required solution is ( − 6 , 8 )

3rd Edition

ISBN: 9780321880116

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

Publisher: PEARSON

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Chapter 4, Problem 26RE

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To solve: The system of linear equations y=−56x+3 and y=−43x by substitution.

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Chapter 4, Problem 25RE

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The system of linear equations y = − 5 6 x + 3 and y = − 4 3 x 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 y=−56x+3 and y=−43x by substitution.

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