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 obtained ordered pair by putting the corresponding value to the variables of the system of linear equations.” Calculation: Consider the given equations. x 2 + y 3 = 1 12 (1) 2 x 3 + y 3 = − 1 3 (2) Step 1: In equation (2), isolate the variable y from x . 2 x 3 + y 3 = − 1 3 y 3 = − 1 3 − 2 x 3 y = − 1 − 2 x Step 2: Substitute y = − 1 − 2 x in the equation (1). x 2 + y 3 = 1 12 x 2 + ( − 1 − 2 x ) 3 = 1 12 Here, the variable y is eliminated. Step 3: Solve the equation obtained in step 2. x 2 + ( − 1 − 2 x ) 3 = 1 12 3 x + 2 ( − 1 − 2 x ) 6 = 1 12 ( Combine like terms ) − x − 2 6 = 1 12 x = − 5 2 Step 4: Now, substitute x = − 5 2 in one of the given equation and obtain the value of y . Replace x = − 5 2 in the equation (2). 2 x 3 + y 3 = − 1 3 2 3 ( − 5 2 ) + y 3 = − 1 3 y 3 = − 1 3 + 5 3 y = 4 Thus, the required solution is ( − 5 2 , 4 )

4th Edition

ISBN: 9780134556079

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

Publisher: Pearson,

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Chapter 4.2, Problem 51E

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To solve: The system of linear equations x2+y3=112 and 2x3+y3=−13 by substitution method.

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Chapter 4.2, Problem 50E

Chapter 4.2, Problem 52E

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Chapter 4 Solutions

Elementary & Intermediate Algebra

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The system of linear equations x 2 + y 3 = 1 12 and 2 x 3 + y 3 = − 1 3 by substitution method.

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

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To solve: The system of linear equations x2+y3=112 and 2x3+y3=−13 by substitution method.

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