# 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

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!

thank you ma'am