Explanation Procedure used: Solving system of linear equations by elimination: (1) If the equations are not in the standard form, they are converted accordingly into a standard form a x + b y = c , where x and y are variables and a , b and c are appropriate constants greater than one. (2) The variable to be eliminated is recognized from the given system of equations. (3) Obtain the LCM (lowest common multiple) of coefficients of the variable to be eliminated. The equations are multiplied with appropriate multipliers so that the coefficients in each equation become additive inverses. Adding the new equations will result in a fresh equation with the decided variable to been eliminated. (4) The equation in step (3) is solved to obtain the value of remaining variable. (5) The value of variable in step (4) is substituted to any of the equation in the given system. (6) Solving the equation obtained in step (5) gives the value of variable eliminated in step (3). (7) The correctness of solution is asserted by substituting back the values in given equations of the system. Calculation: The given system of linear equations is: x + 5 3 y = − 1 (1) 1 2 x + 1 4 y = 1 4 (2) Step 1: Notice that, the given equations are already in standard form. Step 2: The coefficient of x in equation (1) is 1 and in equation (2) is 1 2 . So, x is a better choice of variable to be eliminated. Step 3: Now, multiply equation (2) by 2 and subtract from equation (1) as follows. x + 5 3 y = − 1 x + 1 2 y = 1 2 _ 7 6 y = − 3 2 The obtained equation is, 7 6 y = − 3 2 . Step 4: The equation obtained in Step (3) is solved as, y = ( − 3 2 ) ( 7 6 ) = − 9 7 Thus, the value of y is − 9 7 . Step 5: Substitute y = − 9 7 in the equation (1) as follows. x + 5 3 y = − 1 x + 5 3 ( − 9 7 ) = − 1 x − 15 7 = − 1 x = − 1 + 15 7 Step 6: The equation obtained in Step (5) is solved as follows

3rd Edition

ISBN: 9780321880116

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

Publisher: PEARSON

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Chapter 4.3, Problem 44E

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To solve: The system of equations {x+53y=−112x+14y=14 using elimination.

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Chapter 4.3, Problem 43E

Chapter 4.3, Problem 45E

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The system of equations { x + 5 3 y = − 1 1 2 x + 1 4 y = 1 4 using elimination.

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

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To solve: The system of equations {x+53y=−112x+14y=14 using elimination.

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