Explanation Procedure used: “Solving system of linear equations by elimination” (1) If the equations are not in the standard form, then they are converted accordingly into a standard form which is as follows: a x + b y = c Where x and y are variables and a , b and c are appropriate integer constants. (2) The variable to be eliminated is recognized from the given equations in the system. (3) After obtaining 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 and 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 obtaining value of remaining variable. (5) The value of variable in step (4) is substituted to the equation in any of the given system equations. (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: Name the given equations as, 4 x − 3 y = − 10 (1) − 2 3 x + y = 11 3 (2) Step 1: The equation (1) is already in standard form, no need to arrange in standard form. Transform equation (2) in the standard form a x + b y = c . − 2 3 x + y = 11 3 3 × ( − 2 3 x + y = 11 3 ) − 2 x + 3 y = 11 Step 2: Arbitrarily select variable x for elimination, in the given linear system of equations. Step 3: The coefficients of x in the equations are 4 and − 2 , whose LCM is 4 . Thus, in order to make the coefficients of x to additive inverses, equation (1) is to be multiplied by 1 and equation (2) is to be multiplied by 2 . Since the coefficients of x became additive inverses, the equations can be added as follows. 4 x − 3 y = − 10 − 4 x + 6 y = 22 _ 3 y = 12 Step 4: From step 3, the value of y = 4 . Step 5: Substitute y = 4 in equation (1)

Elementary & Intermediate Algebra with Access Code

3rd Edition

ISBN: 9780321915139

Author: Michael III Sullivan

Publisher: PEARSON

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Chapter 4.3, Problem 39E

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To solve: The system of linear equations 4x−3y=−10 and −23x+y=113 by elimination.

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Chapter 4.3, Problem 38E

Chapter 4.3, Problem 40E

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Answer the number questions with the following answers +/- 2 sqrt(2) +/- i sqrt(6) (-3 +/-3 i sqrt(3))/4 +/-1 +/- sqrt(6) +/- 2/3 sqrt(3) 4 -3 +/- 3 i sqrt(3)

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Chapter 4 Solutions

Elementary & Intermediate Algebra with Access Code

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 4 x − 3 y = − 10 and − 2 3 x + y = 11 3 by elimination.

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To solve: The system of linear equations 4x−3y=−10 and −23x+y=113 by elimination.

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Chapter 4 Solutions