simultaniouse inequalities

[Dak]

:sigh: simultaniouse equasions seem to be causing me quite a lot of grief lately.

 

basically, im relatively adept at solving simultaniouse equasions, but what do i have to be aware of when solving simultaniouse inequalities, ie

 

2x + y < 200

x + 2y < 500

 

and stuff like that. is it even possible to solve simultaniouse inequalities such as this?

 

thanx

[Dak]

to elaborate slightly: with equalities, you can times both sides by -1, no probs, but i believe with inequalities, if you times both sides by -1 you must invert the sighn, ie

 

x > y

 

becomes

 

-x < -y

 

its complications like that that im enquireing about.

 

cheers

[The Thing]

Yes if you divide or multiply each side by a negative you have to invert the sign.

Solving simultaneous inequalities is possible, yes, but it requires graphing of the individual ones and using the feasable area.

[Dak]

how would one do this? i suppose, if you 'solve' an inequality, then you would end up with something like

 

x < y

 

and could re-solve it a different way to get

 

x > z

 

thus

 

z < x < y

 

and keep narrowing the z and y terms down to get a more accurat range, is this akin to the 'graphing'?

[The Thing]

Well, lets get an example.

Say, 2x + y < 200

x + 2y < 500

Well, graph each inequality. To do this you have to solve for y in the inequalities to write it in its slope-intercept form.

y<200-2x

y<(500-x)/2

Alright. Now, if y<mx+c then the inequality is satisfied by all the points BELOW the drawn line and if y>mx+c then its satisfied by those ABOVE the lines. And since we have y<mx+c for both they're satisfied by the points below.

Now to graph them as if they were equations. Kind of hard to visualize here.

I got this picture crudely drawn of what the graph might look like. This is not what it looks like but merely like a model.

So, the attachment. The 2 lines are the graphed inequalities. Now, the black area is what is known as a feasible area (or region). It is basically a region where all the points satisfy the system of inequalities.

Now we can see that our feasible area is a quadrilateral. So we find out the coordinates of each of the vertices of the feasible area.

Which is I don't know.

Now, you list them on paper.

Then, find the one that has no 0s. Now that one is your answer. Now say if the coordinates you got were (0,0), (0, 34), (30,0), and (5,18) for the vertices, the last one would be your answer. So your answer would be x<5, y<18.

Also, as a note, simultaneous inequalities are mainly used in everyday life by businesses to calculate profits and costs. The middle point (the intersecting point of the 2 lines) is known as the optimum point at where the profits in the calculation are maximum and the costs are minimum. It is possible to have more than 1 of these points.If you want I can later put on how simulatenous equations are used to calculate these things. But not going to do now cuz I feel no motivations currently lol.

[Dak]

ah, cool, thank you very much. is there a way to do it without resorting to graph paper, or is it by far easyer to just draw the graph. and what is the prinsiple when you have 3+ inequities? would you take the points at which they crossed over and take the lowest values (if the inequity is of a less-than nature)?

[Dave]

:sigh: simultaniouse equasions seem to be causing me quite a lot of grief lately.

 

basically' date=' im relatively adept at solving simultaniouse equasions, but what do i have to be aware of when solving simultaniouse inequalities, ie

 

2x + y < 200

x + 2y < 500

 

and stuff like that. is it even possible to solve simultaniouse inequalities such as this?

 

thanx[/quote']

 

I find the easiest way of doing this is to convert it all into one equation. You can write this:

 

[math](2x + y) + (x + 2y) < 200 + 500 = 700[/math]

 

So: [math]3x + 3y < 700[/math] and from there it's reasonably easy.

[Dak]

and then take one of the original terms from it? cool! that is truly cunning.

 

what about any other things that cause the inequity sighn to be reversed? i assume its only timesing by a negative?

 

and is there any thing that can be done with two inequities that have opposite inequities (?)

 

what i mean is

 

3x + y < z

x + 3y > z

 

could i just convert it to

 

3x + y < z

-x - 3y < -z

 

and go from there? seems too easy to be true

 

the reason that i ask is because of this post.

[Dave]

Having said my above post, I'm not entirely sure I'm right. My brain is still in Analysis III mode; I'll have a think and get back to you.

[Dak]

it is correct, its basic algebra. as for the inequities, logically:

 

small smaller than big [plus]

small smaller than big [gives]

 

two smalls smaller than two bigs

 

which is correct, so, in conclusion, your post was still cunning

[Dave]

Whilst my answer is certainly true, it's probably not what you're looking for (the graph up from my post is correct). It's been a while since I've plotted inequalities. If you want real cunning, you should check out the following identity (triangle inequality):

 

[math]|x+y| \leq |x| + |y|[/math]

 

I use this everywhere. It's quite a nice exercise to try and prove it.

[davesbird]

what about any other things that cause the inequity sighn to be reversed? i assume its only timesing by a negative?

 

Yes indeed' date=' multiplying by a negative is the only time when the inequality sign should be reversed.

 

What can become tricky is when you have an equation like [math']\frac{1}{x+2} < 4[/math]

 

In this scenario, you must square the bottom, (and indeed the whole equation) before muliplying up, as you don't know whether [math] x-2 [/math] is positive or negative, ie. continue with:

 

[math]\frac{1}{(x+2)^2} < 16[/math]

 

[math]1 < 16(x+2)^2[/math]

[Dak]

[math]|x+y| \leq |x| + |y|[/math']I use this everywhere. It's quite a nice exercise to try and prove it.

can i use logic instead of math?

 

if y = positive then [math]|x+y| = |x| + |y|[/math]

 

if y = negative then [math]|x+y| < |x| + |y|[/math], as it essentially translates to [math]|x-|y|| \leq |x| + |y|[/math]? ie x - a term is less then x + a term (where term is a positive number)

 

therefore, combining the two, gives [math]|x+y| \leq |x| + |y|[/math]?

[Dave]

Not really Plus I'm not quite sure what you're going on about there, for instance - what if x is negative?

 

A nice (and standard) proof goes as follows:

 

[math](|x|+|y|)^2 = |x|^2 + |y|^2 + 2|xy| = x^2 + y^2 + 2|xy|[/math]

 

Also, [math]|x+y|^2 = (x+y)^2 = x^2 + y^2 + 2xy[/math].

 

So [math](|x|+|y|)^2 - |x+y|^2 = 2|xy| - 2xy[/math]

 

This quantity is certainly greater than or equal to zero, since [math]|xy| \geq xy[/math]. Hence:

 

[math](|x|+|y|)^2 - |x+y|^2 \geq 0[/math], delivering the required result.

[Dak]

our definitions of 'nice' obviously differ quite alot. i would consider a nise proof to be:

 

four possibilities:

 

X..Y

-..+

-..-

+..+

+..-

 

if x and y are positive, then |x + y| = |x| + |y|

if x and y are -, then |x + y| = |x| + |y|

which is consistant with the 'or equal to' part of the [math] \leq [/math] assertation

 

if x is + and y is -, then logically speaking, remembering y is negative,

x + y < x, and so |x + y| < |x|, whereas |x| + |y| will obviously be greater than x, as you are adding a positive to it (|y| being positive), hense

|x + y| < |x| + |y| is logically equivelant to

 

something less than x < something more than x

 

which is true. (this is what i was (badly) trying to say when i said |x+|y||<|x|+|y|)

 

ritey dokey, so far weve adressed three of the four possibilities and found them to be consistent with the statement [math]|x+y| \leq |x|+|y|[/math], all thats left is if y is - and x is +: which is identical to the issue of x being + and y being -.

 

so, whatever the values (+ or -), the term |x+y| never exceeds the term |x|+|y|.

 

tada! proven. i think this way is 'nicer' cos it doesnt involve maths, though im sure youd disagree

[Johnny5]

Whilst my answer is certainly true' date=' it's probably not what you're looking for (the graph up from my post is correct). It's been a while since I've plotted inequalities. If you want real cunning, you should check out the following identity (triangle inequality):

 

[math']|x+y| \leq |x| + |y|[/math]

 

I use this everywhere. It's quite a nice exercise to try and prove it.

 

I tried to prove it once too, you learn something about straight lines, and distance.