Answer Letting z = a , where a is any real number, the solutions of the original system are all ordered triples of the form ( 3 − a , 1 + a , a ) Explanation Given information : The system of equation is { 3 x − 3 y + 6 z = 6 x + 2 y − z = 5 5 x − 8 y + 13 z = 7 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 3 x − 3 y + 6 z = 6 ▶ 1 s t equation x + 2 y − z = 5 ▶ 2 n d equation 5 x − 8 y + 13 z = 7 ▶ 3 r d equation In order to eliminate x , multiply -3 with the 2 n d equation and add the result to the 1 s t equation − 3 x − 6 y + 3 z = − 15 3 x − 3 y + 6 z = 6 _ − 9 y + 9 z = − 9 Simplify the equation − 9 y + 9 z = − 9 by dividing -9 throughout equation − 9 y − 9 + 9 z − 9 = − 9 − 9 y − z = 1 Label the new equation as 4 t h equation y − z = 1 ▶ 4 t h equation In order to eliminatex , multiply -5 with the 2 n d equation and add the result the 3 r d equation − 5 x − 10 y + 5 z = − 25 5 x − 8 y + 13 z = 7 _ − 18 y + 18 z = − 18 Label the new equation as 5 t h equation − 18 y + 18 z = − 18 ▶ 5 t h equation Multiply 18 with 4 t h equation and add the result with 5 t h equation 18 y − 18 z = 18 − 18 y + 18 z = − 18 _ 0 = 0 T R U E There are infinitely many solutions to the given system of equation Let z = a , substituting it 4th equation y − a = 1 y − a + a = 1 + a y = 1 + a Substituting z = a and y = 1 + a in 2nd equation and solving for x Distributing 2 in the left side of the equation Combine like terms Subtract 2 and a on both sides of the equation x + 2 ( 1 + a ) − a = 5 x + 2 + 2 a − a = 5 x + 2 + a = 5 x + 2 + a = 5 − 2 − a x = 3 − a Let a be any real number then the solution to the given system of equation is given by the coordinate ( 3 − a , 1 + a , a ) Conclusion: Letting z = a , where a is any real number, the solutions of the original system are all ordered triples of the form ( 3 − a , 1 + a , a )

3rd Edition

ISBN: 9781133947202

Author: Ron Larson

Publisher: Cengage Learning

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Question

Chapter 7.3, Problem 37E

To determine

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

Expert Solution & Answer

Answer to Problem 37E

Letting z = a , where a is any real number, the solutions of the original system are all

ordered triples of the form (3−a, 1+a, a)

Explanation of Solution

Given information : The system of equation is {3x−3y+6z=6x+2y−z=55x−8y+13z=7

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	3x−3y+6z=6	▶ 1^s^t equation
	x+2y−z=5	▶ 2ⁿ^d equation
	5x−8y+13z=7	▶ 3^r^d equation
In order to eliminate x , multiply -3 with the 2ⁿ^d equation and add the result to the 1^s^tequation	−3x−6y+3z=−15 3x−3y+6z=6_ −9y+9z=−9
Simplify the equation −9y+9z=−9 by dividing -9 throughout equation	−9y−9+9z−9=−9−9 y−z=1
Label the new equation as 4^t^h equation	y−z=1	▶ 4^t^h equation
In order to eliminatex , multiply -5 with the 2ⁿ^d equation and add the result the 3^r^d equation	−5x−10y+5z=−25 5x−8y+13z= 7 _ −18y+18z=−18
Label the new equation as 5^t^h equation	−18y+18z=−18	▶ 5^t^h equation
Multiply 18 with 4^t^h equation and add the result with 5^t^h equation	18y−18z=18−18y+18z=−18_ 0=0 TRUE There are infinitely many solutions to the given system of equation
Let z=a , substituting it 4th equation	y−a=1y−a+a=1+ay=1+a
Substituting z=a and y=1+a in 2nd equation and solving for x Distributing 2 in the left side of the equation Combine like terms Subtract 2 and a on both sides of the equation	x+2(1+a)−a=5x+2+2a−a=5x+2+a=5x+2+a=5−2−ax=3−a
Let a be any real number then the solution to the given system of equation is given by the coordinate	(3−a, 1+a, a)

Conclusion:

Letting z = a , where a is any real number, the solutions of the original system are all

ordered triples of the form (3−a, 1+a, a)

Chapter 7.3, Problem 36E

Chapter 7.3, Problem 38E

Chapter 7 Solutions

Precalculus with Limits

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