4.2 Systems of Linear Equations in Three Variables...Title Microsoft Word - CG5_oddSolutions_S42.doc Author Ross Rueger Created Date 10/18/2010 6:05:22 AM

Intermediate Algebra Problem Set 4.2 Solutions to Every Odd-Numbered Problem

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4.2 Systems of Linear Equations in Three Variables 1. Adding the first two equations and the first and third equations results in the system:

2x + 3z = 5

2x ! 2z = 0

Solving the second equation yields x = z, now substituting:

2z + 3z = 5

5z = 5

z = 1

So x = 1, now substituting into the original first equation:

1+ y +1 = 4

y + 2 = 4

y = 2

The solution is (1,2,1). 3. Adding the first two equations and the first and third equations results in the system:

2x + 3z = 13

3x ! 3z = !3

Adding yields:

5x = 10

x = 2

Substituting to find z:

2 2( ) + 3z = 13

4 + 3z = 13

3z = 9

z = 3

Substituting into the original first equation:

2 + y + 3 = 6

y + 5 = 6

y = 1

The solution is (2,1,3).


Name _________________________ Date _________________________




5. Adding the second and third equations: 5x + z = 11 Multiplying the second equation by 2:

x + 2y + z = 3

4x ! 2y + 4z = 12

Adding yields:

5x + 5z = 15

x + z = 3

So the system becomes:

5x + z = 11

x + z = 3

Multiply the second equation by –1:

5x + z = 11

!x ! z = !3

Adding yields:

4x = 8

x = 2


5 2( ) + z = 11

z +10 = 11

z = 1


2 + 2y +1 = 3

2y + 3 = 3

2y = 0

y = 0



Name _________________________ Date _________________________




7. Multiply the second equation by –1 and add it to the first equation:

2x + 3y ! 2z = 4

!x ! 3y + 3z = !4

Adding results in the equation x + z = 0 . Multiply the second equation by 2 and add it to the third equation:

2x + 6y ! 6z = 8

3x ! 6y + z = !3

Adding results in the equation:

5x ! 5z = 5

x ! z = 1

So the system becomes:

x ! z = 1

x + z = 0

Adding yields:

2x = 1

x =1

2


1

2+ z = 0

z = !1

2


21

2

!"#

$%&+ 3y ' 2 '

1

2

!"#

$%&= 4

1+ 3y +1 = 4

3y + 2 = 4

3y = 2

y =2

3

The solution is 1

2,2

3,!1

2

"#$

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.

9. Multiply the first equation by 2 and add it to the second equation:

!2x + 8y ! 6z = 4

2x ! 8y + 6z = 1

Adding yields 0 = 5, which is false. There is no solution (inconsistent system).


Name _________________________ Date _________________________




11. To clear the system of fractions, multiply the first equation by 2 and the second equation by 3:

x ! 2y + 2z = 0

6x + y + 3z = 6

x + y + z = !4

Multiply the third equation by 2 and add it to the first equation:

x ! 2y + 2z = 0

2x + 2y + 2z = !8

Adding yields the equation 3x + 4z = !8 . Multiply the third equation by –1 and add it to the second equation:

6x + y + 3z = 6

!x ! y ! z = 4

Adding yields the equation 5x + 2z = 10 . So the system becomes:

3x + 4z = !8

5x + 2z = 10


3x + 4z = !8

!10x ! 4z = !20

Adding yields:

!7x = !28

x = 4


3 4( ) + 4z = !8

12 + 4z = !8

4z = !20

z = !5

Substituting into the original third equation:

4 + y ! 5 = !4

y !1 = !4

y = !3

The solution is (4,–3,–5). 13. Multiply the first equation by –2 and add it to the third equation:

!4x + 2y + 6z = !2

4x ! 2y ! 6z = 2

Adding yields 0 = 0, which is true. Since there are now less equations than unknowns, there is no unique solution (dependent system).


Name _________________________ Date _________________________




15. Multiply the second equation by 3 and add it to the first equation:

2x ! y + 3z = 4

3x + 6y ! 3z = !9

Adding yields the equation 5x + 5y = !5 , or x + y = !1 . Multiply the second equation by 2 and add it to the third equation:

2x + 4y ! 2z = !6

4x + 3y + 2z = !5

Adding yields the equation 6x + 7y = !11 . So the system becomes:

6x + 7y = !11

x + y = !1


6x + 7y = !11

!6x ! 6y = 6

Adding yields y = –5. Substituting to find x:

6x + 7 !5( ) = !11

6x ! 35 = !11

6x = 24

x = 4


2 4( ) ! !5( ) + 3z = 4

13+ 3z = 4

3z = !9

z = !3

The solution is (4,–5,–3). 17. Adding the second and third equations results in the equation x + y = 9 . Since this is the same as the first equation, there are less equations than unknowns. There is no unique solution (dependent system).


Name _________________________ Date _________________________




19. Adding the second and third equations results in the equation 4x + y = 3 . So the system becomes:

4x + y = 3

2x + y = 2

Multiplying the second equation by –1:

4x + y = 3

!2x ! y = !2

Adding yields:

2x = 1

x =1

2

Substituting to find y:

21

2

!"#

$%&+ y = 2

1+ y = 2

y = 1

Substituting into the original second equation:

1+ z = 3

z = 2

The solution is 1

2,1,2

!"#

$%&

.


Name _________________________ Date _________________________




21. Multiply the third equation by 2 and adding it to the second equation:

6y ! 4z = 1

2x + 4z = 2

Adding yields the equation 2x + 6y = 3 . So the system becomes:

2x ! 3y = 0

2x + 6y = 3

Multiply the first equation by 2:

4x ! 6y = 0

2x + 6y = 3

Adding yields:

6x = 3

x =1

2


21

2

!"#

$%&+ 6y = 3

1+ 6y = 3

6y = 2

y =1

3

Substituting into the original third equation to find z:

1

2+ 2z = 1

2z =1

2

z =1

4

The solution is 1

2,1

3,1

4

!"#

$%&

.


Name _________________________ Date _________________________




23. Multiply the first equation by –2 and add it to the second equation:

!2x ! 2y + 2z = !4

2x + y + 3z = 4

Adding yields !y + 5z = 0 . Multiply the first equation by –1 and add it to the third equation:

!x ! y + z = !2

x ! 2y + 2z = 6

Adding yields !3y + 3z = 4 . So the system becomes:

!y + 5z = 0

!3y + 3z = 4

Multiply the first equation by –3:

3y !15z = 0

!3y + 3z = 4

Adding yields:

!12z = 4

z = !1

3


!3y + 3 !1

3

"#$

%&'= 4

!3y !1 = 4!3y = 5

y = !5

3


x !5

3+1

3= 2

x !4

3= 2

x =10

3

The solution is 10

3,!5

3,!1

3

"#$

%&'

.


Name _________________________ Date _________________________




25. Multiply the third equation by –1 and add it to the first equation:

2x + 3y = !

1

2

!3y ! 2z =3

4

Adding yields the equation 2x ! 2z =1

4. So the system becomes:

2x ! 2z =1

4

4x + 8z = 2

Multiply the first equation by 4:

8x ! 8z = 1

4x + 8z = 2

Adding yields:

12x = 3

x =1

4


41

4

!"#

$%&+ 8z = 2

1+ 8z = 2

8z = 1

z =1

8


21

4

!"#

$%&+ 3y = '

1

2

1

2+ 3y = '

1

2

3y = '1

y = '1

3

The solution is 1

4,!1

3,1

8

"#$

%&'

.


Name _________________________ Date _________________________




27. To clear each equation of fractions, multiply the first equation by 6, the second equation by 4, and the third equation by 12:

2x + 3y ! z = 24

x ! 3y + 2z = 6

6x ! 8y ! 3z = !64

Multiply the first equation by 2 and add it to the second equation:

4x + 6y ! 2z = 48

x ! 3y + 2z = 6

Adding yields the equation 5x + 3y = 54 . Multiply the first equation by –3 and add it to the third equation:

!6x ! 9y + 3z = !72

6x ! 8y ! 3z = !64

Adding yields:

!17y = !136

y = 8

Substituting to find x:

5x + 3 8( ) = 54

5x + 24 = 54

5x = 30

x = 6


6 ! 3 8( ) + 2z = 6

!18 + 2z = 6

2z = 24

z = 12



Name _________________________ Date _________________________




29. To clear each equation of fractions, multiply the first equation by 6, the second equation by 6, and the third equation by 12:

6x ! 3y ! 2z = !8

2x ! 3z = 30

!3x + 8y !12z = !9

Multiply the first equation by 8 and the third equation by 3:

48x ! 24y !16z = !64

!9x + 24y ! 36z = !27

Adding yields the equation 39x ! 52z = !91 , or 3x ! 4z = !7 . So the system becomes:

2x ! 3z = 30

3x ! 4z = !7

Multiply the first equation by 3 and the second equation by –2:

6x ! 9z = 90

!6y + 8z = 14

Adding yields:

!z = 104

z = !104


2x ! 3 !104( ) = 30

2x + 312 = 30

2x = !282

x = !141


6 !141( ) ! 3y ! 2 !104( ) = !8

!846 ! 3y + 208 = !8

!3y ! 638 = !8

!3y = 630

y = !210

The solution is (–141,–210,–104).


Name _________________________ Date _________________________




31. Divide the second equation by 5 and the third equation by 10 to produce the system:

x ! y ! z = 0

x + 4y = 16

2y ! z = 5

Multiply the third equation by –1 and add it to the first equation:

x ! y ! z = 0

!2y + z = !5

Adding yields the equation x ! 3y = !5 . So the system becomes:

x + 4y = 16

x ! 3y = !5


x + 4y = 16

!x + 3y = 5

Adding yields:

7y = 21

y = 3


x +12 = 16

x = 4


6 ! z = 5

z = 1

The currents are 4 amps, 3 amps, and 1 amp. 33. Translating into symbols: 3x + 2

35. Simplifying: 25 ! 3859

=225

9!385

9= !

160

9

37. Simplifying: 0.08 4,000( ) = 320 39. Applying the distributive property: 10 0.2x + 0.5y( ) = 2x + 5y 41. Solving the equation:

x + 3x + 2( ) = 264 x + 2 = 26

4 x = 24

x = 6


Name _________________________ Date _________________________




43. Adding the two equations:

!9y = 9

y = !1

Substituting into the second equation:

!7 !1( ) + 4 z = 277 + 4 z = 27

4 z = 20

z = 5

The solution is (–1,5).

Documents

4.2 Systems of Linear Equations in Three Variables...Title Microsoft Word - CG5_oddSolutions_S42.doc Author Ross Rueger Created Date 10/18/2010 6:05:22 AM