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: 1 14 − x 2 = − y 7 (1) y = 1 2 + 7 x 3 (2) Step 1: Convert the equation (1) in standard form as follows. 1 14 − x 2 = − y 7 − 1 2 x + 1 7 y = 1 14 Convert the equation (2) in standard form as follows. y = 1 2 + 7 x 3 − 7 3 x + y = 1 2 Thus, the set of linear equations in standard form is: − 1 2 x + 1 7 y = 1 14 (3) − 7 3 x + y = 1 2 (4) Step 2: The coefficient of y in equation (3) is 1 7 and that in equation (4) is 1 . So, y is a better choice for the variable to be eliminated in the given system. Step 3: Multiply equation (4) by − 1 7 and add with equation (3). − 1 2 x + 1 7 y = − 1 14 − 1 3 x − 1 7 y = − 1 14 _ − 1 6 x = − 1 7 The equation obtained is − 1 6 x = − 1 7 . Step 4: The equation obtained in Step (3) is solved as, − 1 6 x = − 1 7 x = 6 7 Thus, the value of x is 6 7

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ISBN: 9780134556079

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

Publisher: Pearson,

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Chapter 4, Problem 48RE

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To solve: The system of linear equations 114−x2=−y7 and y=12+7x3.

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Chapter 4, Problem 47RE

Chapter 4, Problem 49RE

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Chapter 4 Solutions

Elementary & Intermediate Algebra

Ch. 4.1 - A _____ is a grouping of two or more linear...Ch. 4.1 - A _____ of a system of equations consists of...Ch. 4.1 - Prob. 3E Ch. 4.1 - In Problems 3 and 4, determine whether each...Ch. 4.1 - Prob. 5E Ch. 4.1 - Prob. 6E Ch. 4.1 - Prob. 7E Ch. 4.1 - In Problem 8, solve the system by graphing. 8....Ch. 4.1 - Prob. 9E Ch. 4.1 - If a system of linear equations in two variables...

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The system of linear equations 1 14 − x 2 = − y 7 and y = 1 2 + 7 x 3 .

Solution Summary: The author explains the procedure of solving system of linear equations by elimination.

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To solve: The system of linear equations 114−x2=−y7 and y=12+7x3.

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