To determine To calculate: The solution of the system of equations x + y + z = 2 , 2 x − y + 2 z = − 2 and 3 x + 2 y + 6 z = 1 . Explanation Given Information: The provided system of equations are, x + y + z = 2 , 2 x − y + 2 z = − 2 and 3 x + 2 y + 6 z = 1 Formula used: Addition or Elimination method to solve a linear system of equations: Step1. Write the equations in the form of a x + b y + c z = d . Step2. Eliminate the one variable by using any two equations. Step3. Write the new equations in the form of a x + b y = c . Step4. Multiply one or both equations by a constant, so that after adding the equations, there will be only one variable. Step5. Add both the equations which will result in equation of one variable. Step6. Compute the equation obtained in step5 and find the value of one variable. Step7. Substitute the value obtained in step6 of one variable in any of the new equation of two variables to obtain other one. Step8. Substitute the value obtained in the step6 and step7 in any of the original equation to obtain the third one. Step10. Write the value of variables in correct order. Step11. Check the solution by putting the calculated value in all equations in the system. Calculation: Consider the provided system of equations, x + y + z = 2 …… ( 1 ) 2 x − y + 2 z = − 2 …… ( 2 ) 3 x + 2 y + 6 z = 1 …… ( 3 ) Use equation ( 1 ) and equation ( 2 ) to eliminate x and z and solve for y : Multiply equation ( 1 ) by − 2 , − 2 x − 2 y − 2 z = − 4 2 x − y + 2 z = − 2 _ − 3 y = − 6 y = 2 Thus, the value of y is 2 . Use equation ( 1 ) and equation ( 3 ) to eliminate x : Multiply equation ( 1 ) by − 3 , − 3 x − 3 y − 3 z = − 6 3 x + 2 y + 6 z = 1 _ − y + 3 z = − 5 Therefore, the new equation will be, − y + 3 z = − 5 …… ( 4 ) To determine z , substitute y = 2 into equation ( 4 ) , − y + 3 z = − 5 − 2 + 3 z = − 5 3 z = − 3 z = − 1 Thus, the value of z is − 1

Intermediate Algebra For College Students (9th Edition)

9th Edition

ISBN: 9780321927354

Author: Allen R. Angel, Dennis Runde

Publisher: PEARSON

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Chapter 4.3, Problem 11MCT

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To calculate: The solution of the system of equations x+y+z=2, 2x−y+2z=−2 and 3x+2y+6z=1.

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Chapter 4.3, Problem 10MCT

Chapter 4.3, Problem 12MCT

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