Math people, need some direction

[HonkeyDonk]

Originally posted by: ariafrost

Originally posted by: HonkeyDonk
by the way, this ain't calc

If you think this is calculus, you ahve some serious problems.

This is like pre-algebra level. Any 6th grader will be able to do this.

Riight... except that that's more of a high-school algebra concept that most people don't get at all... if you're going to argue that it's basic algebra, fine. Then I knew algebra since I was 3! :Q

Ok, maybe 6th is a little early, but I was definitely taught this stuff in 7th grade.

 

[desteffy]

Originally posted by: HonkeyDonk

Originally posted by: desteffy
correct. you can sometimes reduce it to certain bounds for x and y, as you have done above. However this reduces your feasible region, so it isnt something you generally want to do.

what do you mean "reduces your feasible region"

My solution and if you were to graph it would give you the exact same results.

Whatever you graph, those infinite solutions for X and Y would be the exact solutions for what I solved earlier.

reduces your solution set.

actually I looked back up at the problem and your answer is not even correct.

OP just graph the lines and shade in the region of all possible solutions, thats probably what it means.

 

[HonkeyDonk]

Originally posted by: desteffy

Originally posted by: HonkeyDonk

Originally posted by: desteffy
correct. you can sometimes reduce it to certain bounds for x and y, as you have done above. However this reduces your feasible region, so it isnt something you generally want to do.

what do you mean "reduces your feasible region"

My solution and if you were to graph it would give you the exact same results.

Whatever you graph, those infinite solutions for X and Y would be the exact solutions for what I solved earlier.

reduces your solution set.

actually I looked back up at the problem and your answer is not even correct.

OP just graph the lines and shade in the region of all possible solutions, thats probably what it means.

how is it not correct.

Give me an example, like some #'s that would make my solution false.

I love it when people come in and quickly say that something is wrong, but never provide any proof. :roll:

 

[HonkeyDonk]

as long as you put in anything that is greater than -16/6 for Y, then 8x + 2y < 24 will always be true.

 

[HonkeyDonk]

Desteffy, so let's say you plot the two inequalities.

1. Y < 12 - 4x (simplified from 8x+2y < 24)

2. Y > -x + 1

You agree with me that those two inequalities are the correct ones to use right?

Well, if you look at the shaded area, you will notice that the Y value will greater than -16/6 (-2.6667) and that X will not be greater than 3.6667 (or 1+16/6).

So you can say that the solution to the inequalities is:

x < 3.6667
Y > -16/6


I still do not see how my solution is any different than graphing it. I get the exact same solutions.

Unless I used the wrong inequalities, my solution is 100% correct.

 

[desteffy]

ok look at your inequalities:

x < 3.6667
Y > -16/6

for example xy = (0,0) satisfies them, but fails the second inequality of the problem.

x + y > 1

think about what the graph of this would look like when graphed in 2 dimensions

 

[HonkeyDonk]

Originally posted by: desteffy
ok look at your inequalities:

x < 3.6667
Y > -16/6

for example xy = (0,0) satisfies them, but fails the second inequality of the problem.

x + y > 1

think about what the graph of this would look like when graphed in 2 dimensions

Ok, you're right about that.

I was on the stipulation that you first pick a value for Y, say 0, then plug that into the first equation. From there you would get a corresponding X value.

So w/ (0,0), Y > -16/6, so 0 > -16/6 = true.

Then x + 0 > 1, thus X > 1

8(1) + 2(0) < 24 = true.

So in my solution, I'm finding 1 variable value and then plugging that into the other inequalitiy to find the value of the other variable. So it is correct in a sense, depending how you solve for the values.

But I guess you're supposed to be able to just say a set of points (0,0) or something and have it work or not. The shaded region, however, will not be lower than -16/6 for the Y value, and will not be greater than 3.667 for the X value.

 

[zerocool1]

yay simultaneous equations...