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Solving Work Problems Involving System of Linear Equations

Updated on February 28, 2012

Solving Work Problems Involving System of Linear Equations



In this hub I presented several challenging work problems involving system of linear equations complete with solution. I hope this hub will be useful to you and you will enjoy viewing.



Sample Problem Number One :

Two machines are used with production of toys. 1000 toys will be produced in a day if machine A operates for 4 hours and machine B operates for 3 hours or if machine A operates for 6 hours and machine B operates for 2 hours. How long would it take each machine to produce 1000 toys.


Solution :

Let X = Number of hours it will take machine A to produce 1000 toys.

Let Y = Number of hours it will take machine B to operate 1000 toys.

4/X + 3/Y = 1 equation one

6/X + 2/Y = 1 equation two

Multiply equation one by 3 and equation two by -2 to eliminate X and solve for Y:

12/X + 9/Y = 3

-12/X - 4/Y = -2

5/Y = 1 ==è Y = 5 hours


To solve for X you may use substitution method : Using equation one substitute Y = 5

4/X + 3/5 = 1

4/X = 1 - 3/5

4/X = 2/5

2X = 20

X = 10 hours

It will take 10 hours for machine A and 5 hours for machine B to complete 1000 toys each.



Sample Problem Number Two :

Crew #1 and #2 can build a house in 45 days, #2 and #3 for the same job in 36 days, #1 and #3 for the same job in 60 days. If crew #1, #2,#3 work together, how many days can they do the same job ?

Slution:

Let A = number of hours it will take crew #1 to build the house

Let B= number of hours it will take crew#2 to build the house

Let C = number of hours it will take crew #3 to build the house

Let D = number of hours it will take crew #1,#2,#3 to build the house together

45/A + 45/B = 1 equation one

36/B + 36/C = 1 equation two

60/A + 60/C = 1 equation three

D/A + D/B + D/C = 1 equation four

Use equation one and two to elliminate B

(45/A + 45/B = 1) 36

(36/B + 36/C = 1 ) -45

1620/A + 1620/B = 36

-1620/B - 1620/C = -45

1620/A - 1620/C = -9 let this be equation 5

Now use equation 3 and equation 5 to elliminate either A or C

1620/A - 1620/C = -9

(60/A + 60/C = 1) 27

1620/A - 1620/C = -9

1620/A + 1620/C = 27

3240/A = 18

18A = 3240

A = 180 days

To solve for B substitute A = 180 in equation one :

45/180 + 45/B = 1

¼ + 45/B = 1

45/B = 1 – ¼

45/B = ¾

3B = 180

B = 60 days

To solve for C substitute A = 180 in equation three

60/180 + 60/ C = 1

1/3 + 60/C = 1

60/C = 1 – 1/3

60/C = 2/3

180 = 2C

C = 90 days

To solve for D substitute A = 180 B = 60 C = 90 in equation four

(D/180 + D/60 + D/90 = 1) 180

D + 3D + 2D = 180

6D = 180

D = 30 days

It will take 30 days for the three crews to build the house.



Sample Problem Number Three :

A tank is supplied by two pipes A and B and drained by pipe C. IF the tank is full and A and C are opened the tank can be emptied in 10 hours. If B and C are opened the tank can be emptied in 13 hour and 20 minutes. If the tank is empty A and B are opened the tank can be filled in 4 and 4/9 hours. If all pipes are opened , how long it will be filled up.

Solution :

Let W = Number of hours it will take pipe A to fill the tank

Let X = Number of hours it will take pipe B to fill the tank

Let Y = Number of hours it will take pipe C to drain the tank

Let Z = If all pipes are open, the number of hours it will take to fill the tank

10/W - 10/Y = 1 equation one

(40/3 )/X - (40/3)/Y = 1 equation two

(40/9)/W + (40/9)/X = 1 equation three

Z/W + Z/X - Z/Y = 1 equation four

We will solve this using ellimintation method first. Using equation one and equation three to eliminate W:


(10/W - 10/Y = 1 ) multiply by 40/9

((40/9)/W + ( 40/9)/X = 1) multiply by -10

(400/9)/W - (400/9)/Y = 40/9

-(400/9)/w - (400/9)/X = -10

(-400/9)/Y - ( 400/9)/X = -50/9 let this be equation five


Now use equation two and five to eliminate either X or Y

((40/3 )/X - (40/3)/Y = 1) multiply by 400/9

((-400/9)/X - (400/9)/Y = -50/9) multiply by 40/3

(16000/27)/X - (16000/27)/Y = 400/9

(-16000/27)/X - (16000/27)/Y = -2000/27

(-32000/27)/Y = 400/9 - 2000/27

(-32000/27)/Y = (1200 -2000)/27

(-32000/27)/Y = -800/27

800Y = 32000

Y = 40 hours


To solve for W substitute Y = 40 in eqn one

10/W - 10/40 = 1

10/W = 1 + ¼

10/W = 5/4

5W = 40

W = 8 hours

To solve for X substitute Y = 40 in eqn two:

(40/3)/X - (40/3)/40 = 1

40/3X = 1 + 1/3

40/3X = 4/3

4x = 40

X = 10 hours

To solve for Z substitute W = 8, X = 10, Y = 40 in equation four

(Z/8 + Z/10 -Z/40 = 1) multiply by 40

5Z + 4Z - Z = 40

8Z = 40

Z = 5 hours

If pipe A, B and C are open, it will take five hours to fill the tank.





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    • profile image

      mariel 4 years ago

      thank you ma'am

    • cristina327 profile image
      Author

      cristina327 5 years ago from Manila

      Hi teaches12345 I am glad to see you on this Math hub. Thank you for dropping by and appreciating this hub although you are not that mathematically inclined. I am always glad to hear from you. Yes Algebra has sought great applications in the business field and studying algebra is worth investing your time. Blessings to you always. Best regards.

    • teaches12345 profile image

      Dianna Mendez 5 years ago

      My expertise is not math but I can see that those who love the challenge would really benefit from your hub. This would be a great case study for any college course on marketing or operational management. Voted up!