Intermediate Algebra Name Problem Set 4.1 Odd-Numbered … · 2010-11-02 · 8x+6y=28 27x!6y=42 Adding yields: 35x=70 x=2 The solution is (2,2). 19. Multiply the first equation by

Intermediate Algebra Problem Set 4.1 Solutions to Every Odd-Numbered Problem

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4.1 Systems of Linear Equations in Two Variables 1. The intersection point is (4,3). 3. The intersection point is (–5,–6).

5. The intersection point is (4,2).

7. The lines are parallel. There is no solution to the system.


Name _________________________ Date _________________________




9. Solving the two equations:

3x + y = 5

3x ! y = 3

Adding yields:

6x = 8

x =4

3

Substituting into the first equation:

34

3

!"#

$%&+ y = 5

4 + y = 5

y = 1

The solution is 4

3,1

!"#

$%&

.

11. Multiply the first equation by 3:

3x + 6y = 0

2x ! 6y = 5

Adding yields:

5x = 5

x = 1

The solution is 1,!1

2

"#$

%&'

.

13. Multiply the first equation by –2:

!4x +10y = !32

4x ! 3y = 11

Adding yields:

7y = !21

y = !3

The solution is 1

2,!3"

#$%&'

.


Name _________________________ Date _________________________




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

18x + 9y = !3

!18x !10y = !2

Adding yields:

!y = !5

y = 5

The solution is !8

3,5

"#$

%&'

.

17. Multiply the first equation by 2 and the second equation by 3:

8x + 6y = 28

27x ! 6y = 42

Adding yields:

35x = 70

x = 2

The solution is (2,2). 19. Multiply the first equation by 2:

4x !10y = 6

!4x +10y = 3

Adding yields 0 = 9, which is false. There is no solution (! ). 21. To clear each equation of fractions, multiply the first equation by 6 and the second equation by 20:

3x + 2y = 78

8x + 5y = 200

Multiply the first equation by 5 and the second equation by –2:

15x +10y = 390

!16x !10y = !400

Adding yields:

!x = !10

x = 10

The solution is (10,24).


Name _________________________ Date _________________________




23. To clear each equation of fractions, multiply the first equation by 15 and the second equation by 6:

10x + 6y = !60

2x ! 3y = !2

Multiply the second equation by 2:

10x + 6y = !60

4x ! 6y = !4

Adding yields:

14x = !64

x = !32

7

Substituting into the second equation:

2 !32

7

"#$

%&'! 3y = !2

!64

7! 3y = !2

!3y =50

7

y = !50

21

The solution is !32

7,!50

21

"#$

%&'

.

25. Substituting into the first equation: 27. Substituting into the first equation:

7 2y + 9( ) ! y = 24

14y + 63! y = 24

13y = !39

y = !3

6x ! !3

4x !1"

#$%&'= 10

6x +3

4x +1 = 10

27

4x = 9

27x = 36

x =4

3

The solution is (3,–3). The solution is 4

3,!2"

#$%&'

.


Name _________________________ Date _________________________




29. Substituting into the first equation:

4x ! 4 = 3x ! 2

x ! 4 = !2

x = 2

The solution is (2,4). 31. Solving the first equation for y yields y = 2x ! 5 . Substituting into the second equation:

4x ! 2 2x ! 5( ) = 10

4x ! 4x +10 = 10

10 = 10

Since this statement is true, the two lines coincide. The solution is (x, y) | 2x ! y = 5{ } . 33. Substituting into the first equation:

1

3

3

2y

!"#

$%&'1

2y = 0

1

2y '

1

2y = 0

0 = 0

Since this statement is true, the two lines coincide. The solution is (x, y) | x =3

2y

!"#

$%&

.


8x !14y = 6

35x +14y = !21

Adding yields:

43x = !15

x = !15

43

Substituting into the original second equation:

5 !15

43

"#$

%&'+ 2y = !3

!75

43+ 2y = !3

2y = !54

43

y = !27

43

The solution is !15

43,!27

43

"#$

%&'

.


Name _________________________ Date _________________________





27x ! 24y = 12

16x + 24y = 48

Adding yields:

43x = 60

x =60

43


260

43

!"#

$%&+ 3y = 6

120

43+ 3y = 6

3y =138

43

y =46

43

The solution is 60

43,46

43

!"#

$%&

.


6x !10y = 4

35x +10y = 5

Adding yields:

41x = 9

x =9

41


79

41

!"#

$%&+ 2y = 1

63

41+ 2y = 1

2y = '22

41

y = '11

41

The solution is 9

41,!11

41

"#$

%&'

.


Name _________________________ Date _________________________




41. Multiply the second equation by 3:

x ! 3y = 7

6x + 3y = !18

Adding yields:

7x = !11

x = !11

7


2 !11

7

"#$

%&'+ y = !6

!22

7+ y = !6

y = !20

7

The solution is !11

7,!20

7

"#$

%&'

.


!1

3x + 2 =

1

2x +

1

3

6 !1

3x + 2

"#$

%&'= 6

1

2x +

1

3

"#$

%&'

!2x +12 = 3x + 2!5x = !10x = 2

Substituting into the first equation: y =1

22( ) +

1

3= 1+

1

3=4

3. The solution is 2,

4

3

!"#

$%&

.


32

3y ! 4"

#$%&'! 4y = 12

2y !12 ! 4y = 12!2y !12 = 12

!2y = 24y = !12

Substituting into the second equation: x =2

3!12( ) ! 4 = !8 ! 4 = !12 . The solution is (–12,–12).


Name _________________________ Date _________________________




47. Multiply the first equation by 2:

8x ! 6y = !14

!8x + 6y = !11

0 = !25

Since this statement is false, there is no solution (! ). 49. Multiply the first equation by –20:

!60y ! 20z = !340

5y + 20z = 65

Adding yields:

!55y = !275

y = 5


3 5( ) + z = 17

15 + z = 17

z = 2

The solution is y = 5, z = 2. 51. Substitute into the first equation:

3

4x !

1

3

1

4x

"#$

%&'= 1

3

4x !

1

12x = 1

2

3x = 1

x =3

2

Substituting into the second equation: y =1

4

3

2

!"#

$%&=3

8. The solution is

3

2,3

8

!"#

$%&

.


Name _________________________ Date _________________________




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

3x ! 6y = 4

4x ! 3y = !8

Multiply the second equation by –2:

3x ! 6y = 4

!8x + 6y = 16

Adding yields:

!5x = 20

x = !4


3 !4( ) ! 6y = 4

!12 ! 6y = 4

!6y = 16

y = !8

3

The solution is !4,!8

3

"#$

%&'

.

55. a. Simplifying: 3x ! 4 y( ) ! 3 x ! y( ) = 3x ! 4 y ! 3x + 3y = ! y b. Substituting x = 0:

3 0( ) ! 4y = 8

!4y = 8

y = !2

c. From part b, the y-intercept is (0,–2). d. Graphing the line:


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e. Multiply the second equation by –3:

3x ! 4 y = 8!3x + 3y = !6

Adding yields:

! y = 2y = !2


3x ! 4 !2( ) = 83x + 8 = 8

3x = 0

x = 0

The lines intersect at the point (0,–2). 57. Multiply the second equation by 100:

x + y = 10000

6x + 5y = 56000

Multiply the first equation by –5:

!5x ! 5y = !50000

6x + 5y = 56000

Adding yields x = 6000. The solution is (6000,4000). 59. Substituting x = 4 ! y into the second equation:

4 ! y( ) ! 2y = 44 ! 3y = 4

!3y = 0

y = 0

The solution is (4,0). 61. Simplifying: 2 ! 2 6( ) = 2 !12 = !10 63. Simplifying: x + 3y( ) !1 x ! 2z( ) = x + 3y ! x + 2z = 3y + 2z 65. Solving the equation: 67. Solving the equation:

!9y = !9

y = 1

3 1( ) + 2z = 93+ 2z = 9

2z = 6

z = 3

69. Applying the distributive property: 2 5x ! z( ) = 10x ! 2z 71. Applying the distributive property: 3 3x + y ! 2z( ) = 9x + 3y ! 6z

Documents

Intermediate Algebra Name Problem Set 4.1 Odd-Numbered … · 2010-11-02 · 8x+6y=28 27x!6y=42 Adding yields: 35x=70 x=2 The solution is (2,2). 19. Multiply the first equation by