Solving 3 systems of inequalities with 3 variables

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Loganblahtimes2

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"Each week at a furniture factory, there are 2000 work hours available in the construction department, 1400 work hours in the painting department, and 1300 work hours in the packing department. Producing a chair requires 2 hours of construction, 1 hour or painting and 2 hours for packing. Producing a table requires 4 hours of construction, 3 hours of painting and 3 hours for packing. Producing a chest requires 8 hours of construction, 6 hours of painting and 4 hours for packing. If all available time is used in every department, how many of each item are produced each week? Show your system of equation and steps/method used for solution along with your final answer."

I'm having trouble on deciphering this problem. I realize this would be a system of 3 inequalities each with 3 variables. Here is the system of inequalities:
2x+4y+8z<=2000
x+3y+6z<=1400
2x+3y+4x<=1300
x>=0
y>=0
z>=0
(X being chairs, Y being tables, Z being chests.)

In this class, we were never told how to graph an inequality with 3 variables, and this is on a test. I am clueless as to how to start this as I do not know how I would graph the Z variable into the line, much less find the final solution.
This is on a test in a starting level college math class. I wish not to name the course for a little bit of anonymity, but the class does not focus exclusively around probability and statistics. Putting that bit here because of my last thread where I was told to mention what class I was in.

Thank you for any help in advance.
 
Loganblahtimes2 said:
"Each week at a furniture factory, there are 2000 work hours available in the construction department, 1400 work hours in the painting department, and 1300 work hours in the packing department. Producing a chair requires 2 hours of construction, 1 hour or painting and 2 hours for packing. Producing a table requires 4 hours of construction, 3 hours of painting and 3 hours for packing. Producing a chest requires 8 hours of construction, 6 hours of painting and 4 hours for packing. If all available time is used in every department, how many of each item are produced each week? Show your system of equation and steps/method used for solution along with your final answer."

I'm having trouble on deciphering this problem. I realize this would be a system of 3 inequalities each with 3 variables. Here is the system of inequalities:
2x+4y+8z<=2000
x+3y+6z<=1400
2x+3y+4x<=1300
x>=0
y>=0
z>=0
(X being chairs, Y being tables, Z being chests.)

In this class, we were never told how to graph an inequality with 3 variables, and this is on a test. I am clueless as to how to start this as I do not know how I would graph the Z variable into the line, much less find the final solution.
This is on a test in a starting level college math class. I wish not to name the course for a little bit of anonymity, but the class does not focus exclusively around probability and statistics. Putting that bit here because of my last thread where I was told to mention what class I was in.

Thank you for any help in advance.
Click to expand...
In a word problem, first thing you have to do is to "name things"? What are those x, y & z? Define those!!
 
Follow Subhotosh Khan's advice.

Then ask yourself why "if all available time is used in every department hours" why you call this an inequality problem?

Finally, most inequality problems are in many dimensions. You cannot graph them to get a solution. How can you proceed without graphing?
 
JeffM said:
Follow Subhotosh Khan's advice.

Then ask yourself why "if all available time is used in every department hours" why you call this an inequality problem?

Finally, most inequality problems are in many dimensions. You cannot graph them to get a solution. How can you proceed without graphing?
Click to expand...
I have X, Y and Z defined as "X being chairs, Y being tables, Z being chests". I do not know how to proceed, that's why I came here.
Thanks for your patience.
 
Great. x is # chairs, y # tables, and z # chests.

How many hours are available in the construction department? 2000 hours is what is says in the problem, correct. And all of them are used. So

SOMETHING EQUALS 2000. It is not an inequality problem.

In fact, you were 90% there with what you did.

The hours required in construction for each chair are 2. The hours required in construction for each table are 4. The hours required in construction for a chest are 8. So to build x chairs, y tables, and z chests requires

2x + 4y + 8z. You got that! But for some reason you forgot that all the available hours were used.

So [MATH]2x + 4y + 8z = 2000.[/MATH]
You were VERY close. You just missed the clue about all the available hours being used.

Now can you fix the rest of your set-up?

Can you solve that?
 
You have three equations in three unknowns.
Here are the steps to solve this system.

Step 0: If you can divide any equation(s) by some non zero value to make the numbers smaller without introducing fractions then do so. If you have fractions then multiple by some non-zero number to get rid of the fractions.

Step 1: Pick two equations.

Step 2: Pick a variable.

Step 3: Using the two equations in step 1 eliminate the variable you chose in step 2. This gives you one equation in two variables.

Step 4. Using the equation you did NOT pick in step 1 and one of the equations you did pick in step 1 eliminate the variable you chose in step 2. This gives you a 2nd equation in two variables--the same two variables as the equation you obtained from step 3.

Step 5. Using just the two equations in two variables solve this system for the two variables.

Step 6. Pick any of the three original equations. Substitute in the values for the two variables from step 5 and solve for the remaining variable.

If you need help then please post back showing us your work so we know exactly where you are making your mistake.
 
I believe I've come to an answer.
For reference before I explain, I followed Jomo's response, this Khan Academy video (an explanation on how to solve these types of problems), and this website, which had the answer to the problem but no work shown. I am seeking to understand how to do this type of problem so that was merely an answer checker.
I was mainly confused on the = prospect instead of <=. We solved a problem very similar to this a while back in this class and we used <=, so I thought it was used by default for this. I see that is wrong now and thank you for the clarification.

Step 0: I multiplied the equation x+3y+6z=1400 by 2 to get 2x+6y+12z=2800.
I will refer to 2x+4y+8z=2000 as problem #1 for simplicity.
I will refer to 2x+6y+12z=2800 as #2.
I will refer to 2x+3y+4z=1300 as #3.

Step 1: Subtracted problem #3 from #1 to obtain y+4z=700, which I will refer to as problem A for simplicity.

Step 2: Subtracted problem #3 from #2 to obtain 3y+8z=1500, which I will refer to as problem B for simplicity.

Step 3: Multiplied problem A by 2 to obtain 2y+8z=1400, which I subtracted from problem B to obtain y=100.

Step 4: Substituted Y for 100 in problem A and solved, obtaining z=150.

Step 5: Substituted Y for 100 and Z for 150 in problem #3 and solved to obtain x=200.

These 3 answers (x=200, y=100, z=150) match the answers shown on the second site and I believe this is correct. Thank you all for the assistance.
 
Please do not call the equations problems. Just call them equations!
 
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