Answer The solution to the given system of equation is ( 3 , 7 2 , 1 2 ) Explanation Given information : The system of equation is { 2 x − 2 y − 6 z = − 4 − 3 x + 2 y + 6 z = 1 x − y − 5 z = − 3 Concept Involved: Solution of a system of equation is the point which makes both the equation TRUE. Graphically the solution to the system of equation is the point where the two lines meet. Method of Elimination: To use the method of elimination to solve a system of two linear equations in x and y, perform the following steps. 1. Obtain coefficients for x (or y) that differ only in sign by multiplying all terms of one or both equations by suitably chosen constants. 2. Add the equations to eliminate one variable. 3. Solve the equation obtained in Step 2. 4. Back-substitute the value obtained in Step 3 into either of the original equations and solve for the other variable. 5. Check that the solution satisfies each of the original equations. The Number of Solutions of a Linear System For a system of linear equations, exactly one of the following is true. 1. There is exactly one solution. 2. There are infinitely many solutions. 3. There is no solution. Calculation: Description Steps Label the given equations 2 x − 2 y − 6 z = − 4 ▶ 1 s t equation − 3 x + 2 y + 6 z = 1 ▶ 2 n d equation x − y − 5 z = − 3 ▶ 3 r d equation In order to eliminate y, add 1 s t and 2 n d equation 2 x − 2 y − 6 z = − 4 − 3 x + 2 y + 6 z = 1 _ − x = − 3 In order to solve for x Divide by -1 on both sides of the equation Simplifying fraction both sides − x − 1 = − 3 − 1 x = 3 In order to eliminate the same y variable, multiply 2 with the 3 r d equation and add the result to the 2 n d equation 2 x − 2 y − 10 z = − 6 − 3 x + 2 y + 6 z = 1 _ − x − 4 z = − 5 Label the new equation as 4 t h equation − x − 4 z = − 5 ▶ 4 t h equation Substitute 3 for x in the 4 t h equation and solve for z Add 3 on both sides Divide -4 on both sides Simplify fraction on both sides of the equation − 3 − 4 z = − 5 − 3 − 4 z + 3 = − 5 + 3 − 4 z = − 2 − 4 z − 4 = − 2 − 4 z = 1 2 Substitute 1 2 for z and 3 for x in 3rd equation 3 − y − 5 ( 1 2 ) = − 3 3 − y − 5 2 = − 3 − y + 1 2 − 1 2 = − 3 − 1 2 − y = − 7 2 − 1 ( − y ) = − 1 ( − 7 2 ) y = 7 2 Calculation (Continued): Description Steps Checking the solution ( 3 , 7 2 , 1 2 ) by substituting in 1st equation 2 x − 2 y − 6 z = − 4 2 ( 3 ) − 2 ( 7 2 ) − 6 ( 1 2 ) = − 4 6 − 7 − 3 = − 4 − 4 = − 4 T R U E Checking the solution ( 3 , 7 2 , 1 2 ) by substituting in 2nd equation − 3 x + 2 y + 6 z = 1 − 3 ( 3 ) + 2 ( 7 2 ) + 6 ( 1 2 ) = 1 − 9 + 7 + 3 = 1 1 = 1 T R U E Checking the solution ( 3 , 7 2 , 1 2 ) by substituting in 3rdequation x − y − 5 z = − 3 1 ( 3 ) − 1 ( 7 2 ) − 5 ( 1 2 ) = − 3 3 − 7 2 − 5 2 = − 3 3 − 12 2 = − 3 3 − 6 = − 3 − 3 = − 3 T R U E Conclusion: The solution to the given system of equation is ( 3 , 7 2 , 1 2 )

4th Edition

ISBN: 9781337516853

Author: Larson

Publisher: CENGAGE CO

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Question

Chapter 7.3, Problem 34E

To determine

To find : the solution to the given system of linear equations

Expert Solution & Answer

Answer to Problem 34E

The solution to the given system of equation is (3,72,12)

Explanation of Solution

Given information : The system of equation is {2x−2y−6z=−4−3x+2y+6z=1 x−y−5z=−3

Concept Involved:

Solution of a system of equation is the point which makes both the equation TRUE.

Graphically the solution to the system of equation is the point where the two lines meet.

Method of Elimination:

To use the method of elimination to solve a system of two linear equations in x and y, perform the following steps.

1. Obtain coefficients for x (or y) that differ only in sign by multiplying all

terms of one or both equations by suitably chosen constants.

2. Add the equations to eliminate one variable.

3. Solve the equation obtained in Step 2.

4. Back-substitute the value obtained in Step 3 into either of the original

equations and solve for the other variable.

5. Check that the solution satisfies each of the original equations.

The Number of Solutions of a Linear System

For a system of linear equations, exactly one of the following is true.

1. There is exactly one solution.

2. There are infinitely many solutions.

3. There is no solution.

Calculation:

Description	Steps
Label the given equations	2x−2y−6z=−4	▶ 1^s^t equation
	−3x+2y+6z=1	▶ 2ⁿ^d equation
	x−y−5z=−3	▶ 3^r^d equation
In order to eliminate y, add 1^s^t and 2ⁿ^d equation	2x−2y−6z=−4−3x+2y+6z=1_ −x=−3
In order to solve for x Divide by -1 on both sides of the equation Simplifying fraction both sides	−x−1=−3−1 x=3
In order to eliminate the same y variable, multiply 2 with the 3^r^d equation and add the result to the 2ⁿ^dequation	2x−2y−10z=−6−3x+2y+6z=1_ −x−4z=−5
Label the new equation as 4^t^h equation	−x−4z=−5	▶ 4^t^h equation
Substitute 3 for x in the 4^t^h equation and solve for z Add 3 on both sides Divide -4 on both sides Simplify fraction on both sides of the equation	−3−4z=−5−3−4z+3=−5+3−4z=−2−4z−4=−2−4 z=12
Substitute 12 for z and 3 for x in 3rd equation	3−y−5(12)=−33−y−52=−3−y+12−12=−3−12−y=−72−1(−y)=−1(−72)y=72

Calculation (Continued):

Description	Steps
Checking the solution (3,72,12) by substituting in 1st equation 2x−2y−6z=−4	2(3)−2(72)−6(12)=−46−7−3=−4−4=−4 TRUE
Checking the solution (3,72,12) by substituting in 2nd equation −3x+2y+6z=1	−3(3)+2(72)+6(12)=1−9+7+3=11=1 TRUE
Checking the solution (3,72,12) by substituting in 3rdequation x−y−5z=−3	1(3)−1(72)−5(12)=−33−72−52=−33−122=−33−6=−3−3=−3 TRUE

Conclusion:

The solution to the given system of equation is (3,72,12)

Chapter 7.3, Problem 33E

Chapter 7.3, Problem 35E

Chapter 7 Solutions

EBK PRECALCULUS W/LIMITS

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