Solving Work Problems Involving System of Linear Equations

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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Comments 3 comments

teaches12345 profile image

teaches12345 4 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!


cristina327 profile image

cristina327 4 years ago from Manila Author

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.


mariel 3 years ago

thank you ma'am

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