To Answer: To solve the system of linear equations with three variables.

Question

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Answer 1

Question

Chapter 4.4, Problem 28E

To determine

To Answer:

To solve the system of linear equations with three variables.

Expert Solution & Answer

Answer to Problem 28E

Solution:

The solution of given system is (−6,12,−4).

Explanation of Solution

1) Concept:

Step 1: Simplify and put all three equations in the form Ax+By+Cz=D, if needed.

Step 2: Use elimination method to eliminate any one of the variables. So that we get two equations with two unknowns.

Step 3: Solve the remaining system of equations just as solving a system of linear equations with two unknowns.

Step 4: Solve for the third variable.

Step 5: Check whether solved solutions satisfy the given system of equations by substituting the values obtained in the given system of linear equations.

2) Given:

13x−14y+z=−9 →(1)12x−13y−14z=−6 →(2)x−12y−z=−8 →(3)

Calculation:

From all the above three equations, we can observe that the coefficients of variables are in fractions.

To perform calculations easily, we can take L.C.M and make fraction as integers.

(I) Simplify all three equations

For simplifying (1) take L.C.M. By doing so, we get

13x−14y+z=−94x−3y+12z12=−94x−3y+12z=−108 →(1)

By simplifying (2), we get

12x−13y−14z=−66x−4y−3z12=−66x−4y−3z=12×−66x−4y−3z=−72 →(2)

Simplifying (3), we get

x−12y−z=−82x−y−2z2=−82x−y−2z=−16 →(3)

Now the system of linear equations is

4x−3y+12z=−108 →(4)6x−4y−3z=−72 →(5)2x−y−2z=−16 →(6)

(II) We can choose to eliminate y

On multiplying (4) by 4 and (5) by 3 and by subtracting the resultant, we can eliminate y from equation (4) and (5)

Multiplying (3) by 4, we get

4(4x−3y+12z)=−10816x−12y+48z=−432, →(7)

Multiplying (5) by 3, we get

3(6x−4y−3z)=−7218x−12y−9z=−216 →(8)

Subtracting (7) from (8), we get

16x−12y+48z=−43218x−12y−9z=−216−2x+57z=−216 ¯→(9)

By multiplying (6) by 4 and subtracting from (4), we can eliminate y from (5) and (6)

By multiplying (5) by 4, we get

4(2x−y−2z=−16)8x−4y−8z=−64 →(10)

Subtracting (5) from (10), we get

6x−4y−3z=−728x−4y−8z=−64−2x+5z=−8 →(11)¯

Now we have two equations with two unknowns, the two equations are

−2x+57z=−216 →(9)−2x+5z=−8 →(11)

Solve for x and z

Subtracting (9) and (11), we get

−2x+57z=−216−2x+5z=−8 52z=−208z=−20852=−4 ¯

Substitute z=−4 in (11), we get

−2x+5z=−8−2x+5(−4)=−8−2x−20=−8−2x=12x=12−2=−6x=−6

III) Solve for third variable

Substitute x=−6 and z=−4 in (4), we get

4x−3y+12z=−1084(−6)−3y+12(−4)=−108−24−3y−48=−108−3y=−108+48+24−3y=−36y=−36−3=12

IV) Verify:

Substitute x=−6, y=12 and z=−4 in (4), we get

4x−3y+12z=−1084(−6)−3(12)+12(−4)=−108−24−36−48=−108−108=−108L.H.S=R.H.S

Conclusion:

(−6,12,−4) is the solution.

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Answer 2

Question

Chapter 4.4, Problem 28E

To determine

To Answer:

To solve the system of linear equations with three variables.

Expert Solution & Answer

Answer to Problem 28E

Solution:

The solution of given system is (−6,12,−4).

Explanation of Solution

1) Concept:

Step 1: Simplify and put all three equations in the form Ax+By+Cz=D, if needed.

Step 2: Use elimination method to eliminate any one of the variables. So that we get two equations with two unknowns.

Step 3: Solve the remaining system of equations just as solving a system of linear equations with two unknowns.

Step 4: Solve for the third variable.

Step 5: Check whether solved solutions satisfy the given system of equations by substituting the values obtained in the given system of linear equations.

2) Given:

13x−14y+z=−9 →(1)12x−13y−14z=−6 →(2)x−12y−z=−8 →(3)

Calculation:

From all the above three equations, we can observe that the coefficients of variables are in fractions.

To perform calculations easily, we can take L.C.M and make fraction as integers.

(I) Simplify all three equations

For simplifying (1) take L.C.M. By doing so, we get

13x−14y+z=−94x−3y+12z12=−94x−3y+12z=−108 →(1)

By simplifying (2), we get

12x−13y−14z=−66x−4y−3z12=−66x−4y−3z=12×−66x−4y−3z=−72 →(2)

Simplifying (3), we get

x−12y−z=−82x−y−2z2=−82x−y−2z=−16 →(3)

Now the system of linear equations is

4x−3y+12z=−108 →(4)6x−4y−3z=−72 →(5)2x−y−2z=−16 →(6)

(II) We can choose to eliminate y

On multiplying (4) by 4 and (5) by 3 and by subtracting the resultant, we can eliminate y from equation (4) and (5)

Multiplying (3) by 4, we get

4(4x−3y+12z)=−10816x−12y+48z=−432, →(7)

Multiplying (5) by 3, we get

3(6x−4y−3z)=−7218x−12y−9z=−216 →(8)

Subtracting (7) from (8), we get

16x−12y+48z=−43218x−12y−9z=−216−2x+57z=−216 ¯→(9)

By multiplying (6) by 4 and subtracting from (4), we can eliminate y from (5) and (6)

By multiplying (5) by 4, we get

4(2x−y−2z=−16)8x−4y−8z=−64 →(10)

Subtracting (5) from (10), we get

6x−4y−3z=−728x−4y−8z=−64−2x+5z=−8 →(11)¯

Now we have two equations with two unknowns, the two equations are

−2x+57z=−216 →(9)−2x+5z=−8 →(11)

Solve for x and z

Subtracting (9) and (11), we get

−2x+57z=−216−2x+5z=−8 52z=−208z=−20852=−4 ¯

Substitute z=−4 in (11), we get

−2x+5z=−8−2x+5(−4)=−8−2x−20=−8−2x=12x=12−2=−6x=−6

III) Solve for third variable

Substitute x=−6 and z=−4 in (4), we get

4x−3y+12z=−1084(−6)−3y+12(−4)=−108−24−3y−48=−108−3y=−108+48+24−3y=−36y=−36−3=12

IV) Verify:

Substitute x=−6, y=12 and z=−4 in (4), we get

4x−3y+12z=−1084(−6)−3(12)+12(−4)=−108−24−36−48=−108−108=−108L.H.S=R.H.S

Conclusion:

(−6,12,−4) is the solution.

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Answer 3

Question

Chapter 4.4, Problem 28E

To determine

To Answer:

To solve the system of linear equations with three variables.

Expert Solution & Answer

Answer to Problem 28E

Solution:

The solution of given system is (−6,12,−4).

Explanation of Solution

1) Concept:

Step 1: Simplify and put all three equations in the form Ax+By+Cz=D, if needed.

Step 2: Use elimination method to eliminate any one of the variables. So that we get two equations with two unknowns.

Step 3: Solve the remaining system of equations just as solving a system of linear equations with two unknowns.

Step 4: Solve for the third variable.

Step 5: Check whether solved solutions satisfy the given system of equations by substituting the values obtained in the given system of linear equations.

2) Given:

13x−14y+z=−9 →(1)12x−13y−14z=−6 →(2)x−12y−z=−8 →(3)

Calculation:

From all the above three equations, we can observe that the coefficients of variables are in fractions.

To perform calculations easily, we can take L.C.M and make fraction as integers.

(I) Simplify all three equations

For simplifying (1) take L.C.M. By doing so, we get

13x−14y+z=−94x−3y+12z12=−94x−3y+12z=−108 →(1)

By simplifying (2), we get

12x−13y−14z=−66x−4y−3z12=−66x−4y−3z=12×−66x−4y−3z=−72 →(2)

Simplifying (3), we get

x−12y−z=−82x−y−2z2=−82x−y−2z=−16 →(3)

Now the system of linear equations is

4x−3y+12z=−108 →(4)6x−4y−3z=−72 →(5)2x−y−2z=−16 →(6)

(II) We can choose to eliminate y

On multiplying (4) by 4 and (5) by 3 and by subtracting the resultant, we can eliminate y from equation (4) and (5)

Multiplying (3) by 4, we get

4(4x−3y+12z)=−10816x−12y+48z=−432, →(7)

Multiplying (5) by 3, we get

3(6x−4y−3z)=−7218x−12y−9z=−216 →(8)

Subtracting (7) from (8), we get

16x−12y+48z=−43218x−12y−9z=−216−2x+57z=−216 ¯→(9)

By multiplying (6) by 4 and subtracting from (4), we can eliminate y from (5) and (6)

By multiplying (5) by 4, we get

4(2x−y−2z=−16)8x−4y−8z=−64 →(10)

Subtracting (5) from (10), we get

6x−4y−3z=−728x−4y−8z=−64−2x+5z=−8 →(11)¯

Now we have two equations with two unknowns, the two equations are

−2x+57z=−216 →(9)−2x+5z=−8 →(11)

Solve for x and z

Subtracting (9) and (11), we get

−2x+57z=−216−2x+5z=−8 52z=−208z=−20852=−4 ¯

Substitute z=−4 in (11), we get

−2x+5z=−8−2x+5(−4)=−8−2x−20=−8−2x=12x=12−2=−6x=−6

III) Solve for third variable

Substitute x=−6 and z=−4 in (4), we get

4x−3y+12z=−1084(−6)−3y+12(−4)=−108−24−3y−48=−108−3y=−108+48+24−3y=−36y=−36−3=12

IV) Verify:

Substitute x=−6, y=12 and z=−4 in (4), we get

4x−3y+12z=−1084(−6)−3(12)+12(−4)=−108−24−36−48=−108−108=−108L.H.S=R.H.S

Conclusion:

(−6,12,−4) is the solution.

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Answer 4

Question

Chapter 4.4, Problem 28E

To determine

To Answer:

To solve the system of linear equations with three variables.

Expert Solution & Answer

Answer to Problem 28E

Solution:

The solution of given system is (−6,12,−4).

Explanation of Solution

1) Concept:

Step 1: Simplify and put all three equations in the form Ax+By+Cz=D, if needed.

Step 2: Use elimination method to eliminate any one of the variables. So that we get two equations with two unknowns.

Step 3: Solve the remaining system of equations just as solving a system of linear equations with two unknowns.

Step 4: Solve for the third variable.

Step 5: Check whether solved solutions satisfy the given system of equations by substituting the values obtained in the given system of linear equations.

2) Given:

13x−14y+z=−9 →(1)12x−13y−14z=−6 →(2)x−12y−z=−8 →(3)

Calculation:

From all the above three equations, we can observe that the coefficients of variables are in fractions.

To perform calculations easily, we can take L.C.M and make fraction as integers.

(I) Simplify all three equations

For simplifying (1) take L.C.M. By doing so, we get

13x−14y+z=−94x−3y+12z12=−94x−3y+12z=−108 →(1)

By simplifying (2), we get

12x−13y−14z=−66x−4y−3z12=−66x−4y−3z=12×−66x−4y−3z=−72 →(2)

Simplifying (3), we get

x−12y−z=−82x−y−2z2=−82x−y−2z=−16 →(3)

Now the system of linear equations is

4x−3y+12z=−108 →(4)6x−4y−3z=−72 →(5)2x−y−2z=−16 →(6)

(II) We can choose to eliminate y

On multiplying (4) by 4 and (5) by 3 and by subtracting the resultant, we can eliminate y from equation (4) and (5)

Multiplying (3) by 4, we get

4(4x−3y+12z)=−10816x−12y+48z=−432, →(7)

Multiplying (5) by 3, we get

3(6x−4y−3z)=−7218x−12y−9z=−216 →(8)

Subtracting (7) from (8), we get

16x−12y+48z=−43218x−12y−9z=−216−2x+57z=−216 ¯→(9)

By multiplying (6) by 4 and subtracting from (4), we can eliminate y from (5) and (6)

By multiplying (5) by 4, we get

4(2x−y−2z=−16)8x−4y−8z=−64 →(10)

Subtracting (5) from (10), we get

6x−4y−3z=−728x−4y−8z=−64−2x+5z=−8 →(11)¯

Now we have two equations with two unknowns, the two equations are

−2x+57z=−216 →(9)−2x+5z=−8 →(11)

Solve for x and z

Subtracting (9) and (11), we get

−2x+57z=−216−2x+5z=−8 52z=−208z=−20852=−4 ¯

Substitute z=−4 in (11), we get

−2x+5z=−8−2x+5(−4)=−8−2x−20=−8−2x=12x=12−2=−6x=−6

III) Solve for third variable

Substitute x=−6 and z=−4 in (4), we get

4x−3y+12z=−1084(−6)−3y+12(−4)=−108−24−3y−48=−108−3y=−108+48+24−3y=−36y=−36−3=12

IV) Verify:

Substitute x=−6, y=12 and z=−4 in (4), we get

4x−3y+12z=−1084(−6)−3(12)+12(−4)=−108−24−36−48=−108−108=−108L.H.S=R.H.S

Conclusion:

(−6,12,−4) is the solution.

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Answer 5

Chapter 4.4, Problem 28E

To determine

To Answer:

To solve the system of linear equations with three variables.

Expert Solution & Answer

Answer to Problem 28E

Solution:

The solution of given system is (−6,12,−4).

Explanation of Solution

1) Concept:

Step 1: Simplify and put all three equations in the form Ax+By+Cz=D, if needed.

Step 2: Use elimination method to eliminate any one of the variables. So that we get two equations with two unknowns.

Step 3: Solve the remaining system of equations just as solving a system of linear equations with two unknowns.

Step 4: Solve for the third variable.

Step 5: Check whether solved solutions satisfy the given system of equations by substituting the values obtained in the given system of linear equations.

2) Given:

13x−14y+z=−9 →(1)12x−13y−14z=−6 →(2)x−12y−z=−8 →(3)

Calculation:

From all the above three equations, we can observe that the coefficients of variables are in fractions.

To perform calculations easily, we can take L.C.M and make fraction as integers.

(I) Simplify all three equations

For simplifying (1) take L.C.M. By doing so, we get

13x−14y+z=−94x−3y+12z12=−94x−3y+12z=−108 →(1)

By simplifying (2), we get

12x−13y−14z=−66x−4y−3z12=−66x−4y−3z=12×−66x−4y−3z=−72 →(2)

Simplifying (3), we get

x−12y−z=−82x−y−2z2=−82x−y−2z=−16 →(3)

Now the system of linear equations is

4x−3y+12z=−108 →(4)6x−4y−3z=−72 →(5)2x−y−2z=−16 →(6)

(II) We can choose to eliminate y

On multiplying (4) by 4 and (5) by 3 and by subtracting the resultant, we can eliminate y from equation (4) and (5)

Multiplying (3) by 4, we get

4(4x−3y+12z)=−10816x−12y+48z=−432, →(7)

Multiplying (5) by 3, we get

3(6x−4y−3z)=−7218x−12y−9z=−216 →(8)

Subtracting (7) from (8), we get

16x−12y+48z=−43218x−12y−9z=−216−2x+57z=−216 ¯→(9)

By multiplying (6) by 4 and subtracting from (4), we can eliminate y from (5) and (6)

By multiplying (5) by 4, we get

4(2x−y−2z=−16)8x−4y−8z=−64 →(10)

Subtracting (5) from (10), we get

6x−4y−3z=−728x−4y−8z=−64−2x+5z=−8 →(11)¯

Now we have two equations with two unknowns, the two equations are

−2x+57z=−216 →(9)−2x+5z=−8 →(11)

Solve for x and z

Subtracting (9) and (11), we get

−2x+57z=−216−2x+5z=−8 52z=−208z=−20852=−4 ¯

Substitute z=−4 in (11), we get

−2x+5z=−8−2x+5(−4)=−8−2x−20=−8−2x=12x=12−2=−6x=−6

III) Solve for third variable

Substitute x=−6 and z=−4 in (4), we get

4x−3y+12z=−1084(−6)−3y+12(−4)=−108−24−3y−48=−108−3y=−108+48+24−3y=−36y=−36−3=12

IV) Verify:

Substitute x=−6, y=12 and z=−4 in (4), we get

4x−3y+12z=−1084(−6)−3(12)+12(−4)=−108−24−36−48=−108−108=−108L.H.S=R.H.S

Conclusion:

(−6,12,−4) is the solution.

Answer 6

Chapter 4.4, Problem 28E

To determine

To Answer:

To solve the system of linear equations with three variables.

Expert Solution & Answer

Answer to Problem 28E

Solution:

The solution of given system is (−6,12,−4).

Explanation of Solution

1) Concept:

Step 1: Simplify and put all three equations in the form Ax+By+Cz=D, if needed.

Step 2: Use elimination method to eliminate any one of the variables. So that we get two equations with two unknowns.

Step 3: Solve the remaining system of equations just as solving a system of linear equations with two unknowns.

Step 4: Solve for the third variable.

Step 5: Check whether solved solutions satisfy the given system of equations by substituting the values obtained in the given system of linear equations.

2) Given:

13x−14y+z=−9 →(1)12x−13y−14z=−6 →(2)x−12y−z=−8 →(3)

Calculation:

From all the above three equations, we can observe that the coefficients of variables are in fractions.

To perform calculations easily, we can take L.C.M and make fraction as integers.

(I) Simplify all three equations

For simplifying (1) take L.C.M. By doing so, we get

13x−14y+z=−94x−3y+12z12=−94x−3y+12z=−108 →(1)

By simplifying (2), we get

12x−13y−14z=−66x−4y−3z12=−66x−4y−3z=12×−66x−4y−3z=−72 →(2)

Simplifying (3), we get

x−12y−z=−82x−y−2z2=−82x−y−2z=−16 →(3)

Now the system of linear equations is

4x−3y+12z=−108 →(4)6x−4y−3z=−72 →(5)2x−y−2z=−16 →(6)

(II) We can choose to eliminate y

On multiplying (4) by 4 and (5) by 3 and by subtracting the resultant, we can eliminate y from equation (4) and (5)

Multiplying (3) by 4, we get

4(4x−3y+12z)=−10816x−12y+48z=−432, →(7)

Multiplying (5) by 3, we get

3(6x−4y−3z)=−7218x−12y−9z=−216 →(8)

Subtracting (7) from (8), we get

16x−12y+48z=−43218x−12y−9z=−216−2x+57z=−216 ¯→(9)

By multiplying (6) by 4 and subtracting from (4), we can eliminate y from (5) and (6)

By multiplying (5) by 4, we get

4(2x−y−2z=−16)8x−4y−8z=−64 →(10)

Subtracting (5) from (10), we get

6x−4y−3z=−728x−4y−8z=−64−2x+5z=−8 →(11)¯

Now we have two equations with two unknowns, the two equations are

−2x+57z=−216 →(9)−2x+5z=−8 →(11)

Solve for x and z

Subtracting (9) and (11), we get

−2x+57z=−216−2x+5z=−8 52z=−208z=−20852=−4 ¯

Substitute z=−4 in (11), we get

−2x+5z=−8−2x+5(−4)=−8−2x−20=−8−2x=12x=12−2=−6x=−6

III) Solve for third variable

Substitute x=−6 and z=−4 in (4), we get

4x−3y+12z=−1084(−6)−3y+12(−4)=−108−24−3y−48=−108−3y=−108+48+24−3y=−36y=−36−3=12

IV) Verify:

Substitute x=−6, y=12 and z=−4 in (4), we get

4x−3y+12z=−1084(−6)−3(12)+12(−4)=−108−24−36−48=−108−108=−108L.H.S=R.H.S

Conclusion:

(−6,12,−4) is the solution.

Answer 7

Chapter 4.4, Problem 28E

To determine

To Answer:

To solve the system of linear equations with three variables.

Answer 8

Expert Solution & Answer

Answer to Problem 28E

Solution:

The solution of given system is (−6,12,−4).

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

1) Concept:

Step 1: Simplify and put all three equations in the form Ax+By+Cz=D, if needed.

Step 2: Use elimination method to eliminate any one of the variables. So that we get two equations with two unknowns.

Step 3: Solve the remaining system of equations just as solving a system of linear equations with two unknowns.

Step 4: Solve for the third variable.

Step 5: Check whether solved solutions satisfy the given system of equations by substituting the values obtained in the given system of linear equations.

2) Given:

13x−14y+z=−9 →(1)12x−13y−14z=−6 →(2)x−12y−z=−8 →(3)

Calculation:

From all the above three equations, we can observe that the coefficients of variables are in fractions.

To perform calculations easily, we can take L.C.M and make fraction as integers.

(I) Simplify all three equations

For simplifying (1) take L.C.M. By doing so, we get

13x−14y+z=−94x−3y+12z12=−94x−3y+12z=−108 →(1)

By simplifying (2), we get

12x−13y−14z=−66x−4y−3z12=−66x−4y−3z=12×−66x−4y−3z=−72 →(2)

Simplifying (3), we get

x−12y−z=−82x−y−2z2=−82x−y−2z=−16 →(3)

Now the system of linear equations is

4x−3y+12z=−108 →(4)6x−4y−3z=−72 →(5)2x−y−2z=−16 →(6)

(II) We can choose to eliminate y

On multiplying (4) by 4 and (5) by 3 and by subtracting the resultant, we can eliminate y from equation (4) and (5)

Multiplying (3) by 4, we get

4(4x−3y+12z)=−10816x−12y+48z=−432, →(7)

Multiplying (5) by 3, we get

3(6x−4y−3z)=−7218x−12y−9z=−216 →(8)

Subtracting (7) from (8), we get

16x−12y+48z=−43218x−12y−9z=−216−2x+57z=−216 ¯→(9)

By multiplying (6) by 4 and subtracting from (4), we can eliminate y from (5) and (6)

By multiplying (5) by 4, we get

4(2x−y−2z=−16)8x−4y−8z=−64 →(10)

Subtracting (5) from (10), we get

6x−4y−3z=−728x−4y−8z=−64−2x+5z=−8 →(11)¯

Now we have two equations with two unknowns, the two equations are

−2x+57z=−216 →(9)−2x+5z=−8 →(11)

Solve for x and z

Subtracting (9) and (11), we get

−2x+57z=−216−2x+5z=−8 52z=−208z=−20852=−4 ¯

Substitute z=−4 in (11), we get

−2x+5z=−8−2x+5(−4)=−8−2x−20=−8−2x=12x=12−2=−6x=−6

III) Solve for third variable

Substitute x=−6 and z=−4 in (4), we get

4x−3y+12z=−1084(−6)−3y+12(−4)=−108−24−3y−48=−108−3y=−108+48+24−3y=−36y=−36−3=12

IV) Verify:

Substitute x=−6, y=12 and z=−4 in (4), we get

4x−3y+12z=−1084(−6)−3(12)+12(−4)=−108−24−36−48=−108−108=−108L.H.S=R.H.S

Conclusion:

(−6,12,−4) is the solution.

Answer 12

Ch. 4.1 - Prob. 1G Ch. 4.1 - Prob. 2G Ch. 4.1 - Prob. 3G Ch. 4.1 - Prob. 4G Ch. 4.1 - Prob. 1P Ch. 4.1 - Prob. 2P Ch. 4.1 - Prob. 3P Ch. 4.1 - Prob. 4P Ch. 4.1 - Prob. 5P Ch. 4.1 - Prob. 6P

To Answer: To solve the system of linear equations with three variables.

To Answer: To solve the system of linear equations with three variables.

Answer to Problem 28E

Explanation of Solution

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To Answer:

To solve the system of linear equations with three variables.