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) Back-substitute the value found in step 3 into one of the original equations. Simplify and find the value of the remaining variable. (5) Check the proposed solution in both of the system’s given equations”. Calculation: Consider the given equations. 6 x − 5 y = 1 (1) 8 x − 2 y = − 22 (2) Step 1: Convert the equation (2) in the form of y in terms of x , 8 x − 2 y = − 22 2 y = 8 x + 22 y = 4 x + 11 Step 2: Substitute 4 x + 11 for y in the equation (1), 6 x − 5 ( 4 x + 11 ) = 1 6 x − 20 x − 55 = 1 Here, the variable y is eliminated. Step 3: Solve the equation obtained in step 2. 6 x − 20 x − 55 = 1 ( Combine like terms ) − 14 x = 1 + 55 x = − 56 14 x = − 4 Step 4: Now, substitute − 4 for x in one of the given equation and obtain the value of y . Replace x with 0.2 in the equation (1). 6 x − 5 y = 1 6 × ( − 4 ) − 5 y = 1 − 5 y = 24 + 1 − 5 y = 25 y can be further solved as − 5 y = 25 y = − 25 5 y = − 5 Thus, the required solution is ( − 4 , − 5 )

4th Edition

ISBN: 9780134556079

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

Publisher: Pearson,

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Chapter 4.3, Problem 63E

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To solve: The system of linear equations 6x−5y=1 and 8x−2y=−22 by substitution method.

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Chapter 4.3, Problem 62E

Chapter 4.3, Problem 64E

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The system of linear equations 6 x − 5 y = 1 and 8 x − 2 y = − 22 by substitution method.

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To solve: The system of linear equations 6x−5y=1 and 8x−2y=−22 by substitution method.

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