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.
