Math people, need some direction

[AgentZap]

How do I solve something like

8x + 2y < 24
x + y > 1

I'm rusty and its been awhile since i learned this in calc. I am having trouble googling this. Does someone have a link where it teaches you or can someone show me an example?

 

[Heisenberg]

Solve for one variable, substitute it into the other equation.

 

[HonkeyDonk]

Originally posted by: AgentZap
How do I solve something like

8x + 2y < 24
x + y > 1

I'm rusty and its been awhile since i learned this in calc. I am having trouble googling this. Does someone have a link where it teaches you or can someone show me an example?

Solve for x: x > 1 - y

Substitute x into 1st eqn: 8(1-y) + 2y < 24

Solve the eqn for y: 8-8y+2y<24

-6y<16

y>-16/6

Use y to solve for x: x > 1- y (y is whatever you chose so that it is > than -16/6)

 

[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.

 

[five40]

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.

agreed, that is junior high math.

 

[RbSX]

Originally posted by: AgentZap
How do I solve something like

8x + 2y < 24
x + y > 1

I'm rusty and its been awhile since i learned this in calc. I am having trouble googling this. Does someone have a link where it teaches you or can someone show me an example?

2(4x+y) < 24

4x+Y < 12

Y < 12 - 4X

8X + 2(12 - 4x) < 24

 

[RbSX]

Originally posted by: HonkeyDonk

Originally posted by: AgentZap
How do I solve something like

8x + 2y < 24
x + y > 1

I'm rusty and its been awhile since i learned this in calc. I am having trouble googling this. Does someone have a link where it teaches you or can someone show me an example?

Solve for x: x > 1 - y

Substitute x into 1st eqn: 8(1-y) + 2y < 24

Solve the eqn for y: 8-8y+2y<24

-6y<16

y>16/6

Use y to solve for x: x > 1- 16/6

sorry bro, thats wrong

 

[AgentZap]

Thanks that is what I needed to kickstart me. I know it isn't calc, but we used it constantly in calc which is the last time I had to use it until this quantitative methods class 🙂

 

[AgentZap]

Originally posted by: RyanSengara

Originally posted by: AgentZap
How do I solve something like

8x + 2y < 24
x + y > 1

I'm rusty and its been awhile since i learned this in calc. I am having trouble googling this. Does someone have a link where it teaches you or can someone show me an example?

2(4x+y) < 24

4x+Y < 12

Y < 12 - 4X

8X + 2(12 - 4x) < 24

Thanks Ryan

 

[HonkeyDonk]

Originally posted by: RyanSengara

Originally posted by: HonkeyDonk

Originally posted by: AgentZap
How do I solve something like

8x + 2y < 24
x + y > 1

I'm rusty and its been awhile since i learned this in calc. I am having trouble googling this. Does someone have a link where it teaches you or can someone show me an example?

Solve for x: x > 1 - y

Substitute x into 1st eqn: 8(1-y) + 2y < 24

Solve the eqn for y: 8-8y+2y<24

-6y<16

y>16/6

Use y to solve for x: x > 1- 16/6

sorry bro, thats wrong

oops, forgot a - on the y>16/6...it's fixed now.

should be the right answer now.

 

[HonkeyDonk]

Originally posted by: RyanSengara

Originally posted by: AgentZap
How do I solve something like

8x + 2y < 24
x + y > 1

I'm rusty and its been awhile since i learned this in calc. I am having trouble googling this. Does someone have a link where it teaches you or can someone show me an example?

2(4x+y) < 24

4x+Y < 12

Y < 12 - 4X

8X + 2(12 - 4x) < 24

actually, you're wrong here bro.

All you did was use the same eqn (the first one), rearragned it, and plugged it back into itself. That doesn't show you anything.

If you take your eqn, 8x + 2(12-4x) < 24 and factor it out you get:

8x + 24 -8x < 24, then
24 < 24 is WRONG. How can 24 be less than 24?

What you should do is, once you get teh Y < 12 - 4X, plug that into the x+y > 1. Then you can see what X is.

X + (12-4x) > 1, -3x > -11, X > 11/3.

 

[desteffy]

What do you mean by "solve" anyways.

That right there is a region. Just given a few inequalities like that it is possible that some are redundant, but not necessairly. Something like this can not generally be simplified down to saying x< some number and y > some number or whatever.

 

[Tab]

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.

Uh not its not I've never seen anything like that before ever, and I am in a GTA Class.

 

[HonkeyDonk]

Originally posted by: desteffy
What do you mean by "solve" anyways.

That right there is a region. Just given a few inequalities like that it is possible that some are redundant, but not necessairly. Something like this can not generally be simplified down to saying x< some number and y > some number or whatever.

er...that's exactly what solve is supposed to mean.

When you are given two inequalities w/ 2 variables like in the OP's example, u can solve it so that you know X is greater than "a certain number", same with y.

You are correct about the "regions" and stuf, b/c that what it is when dealing w/ inequalities, but again, solving those inequalities means finding what the boundaries are i guess.

 

[cirthix]

thats pre algebra lol. calc adds a lot of 'fun' stuff in there like absolute values, powers, random other crap they feel liek throwing at you

 

[Goosemaster]

Originally posted by: cirthix
thats pre algebra lol. calc adds a lot of 'fun' stuff in there like absolute values, powers, random other crap they feel liek throwing at you

Not to mention the onslaught of flesh-eating limits with no limit when it comes to creating a personal hell for each and everyone of us:|

 

[desteffy]

Originally posted by: HonkeyDonk

Originally posted by: desteffy
What do you mean by "solve" anyways.

That right there is a region. Just given a few inequalities like that it is possible that some are redundant, but not necessairly. Something like this can not generally be simplified down to saying x< some number and y > some number or whatever.

er...that's exactly what solve is supposed to mean.

When you are given two inequalities w/ 2 variables like in the OP's example, u can solve it so that you know X is greater than "a certain number", same with y.

You are correct about the "regions" and stuf, b/c that what it is when dealing w/ inequalities, but again, solving those inequalities means finding what the boundaries are i guess.

but then it wouldtn be called solving it. You may ask how to find an upper or lower bound for x or y, but saying that you can solve it like any other system:
ie

[system of equations] <=> x<number, y>number

is not correct. It just cant be done, just think of what the graph of it in the plane looks like.

 

[HonkeyDonk]

Originally posted by: desteffy

Originally posted by: HonkeyDonk

Originally posted by: desteffy
What do you mean by "solve" anyways.

That right there is a region. Just given a few inequalities like that it is possible that some are redundant, but not necessairly. Something like this can not generally be simplified down to saying x< some number and y > some number or whatever.

er...that's exactly what solve is supposed to mean.

When you are given two inequalities w/ 2 variables like in the OP's example, u can solve it so that you know X is greater than "a certain number", same with y.

You are correct about the "regions" and stuf, b/c that what it is when dealing w/ inequalities, but again, solving those inequalities means finding what the boundaries are i guess.

but then it wouldtn be called solving it. You may ask how to find an upper or lower bound for x or y, but saying that you can solve it like any other system:
ie

[system of equations] <=> x<number, y>number

is not correct. It just cant be done, just think of what the graph of it in the plane looks like.

Go read any math book involving inequalities (usually pre alegbra/algebra books) and they will say "solve this system of inequalities"

The world solve isn't strictly tied down to only finding a certain number "x" or "y", etc.

You can solve this set of inequalities by finding out what X is > or < (same goes for y). Simple as that.

 

[DrPizza]

desteffy is correct.

First of all, this isn't 6th grade math. It's typically taught in 8th or 9th grade. You were close, Honkeydonk. But, your solution indicates that you're in 8th grade (or perhaps 5th grade if it IS taught in 6th grade) because your solution is completely incorrect.

The solution isn't a particular value - the solution is an infinitely large set of points as can be simply depicted by a graph.

 

[desteffy]

Originally posted by: HonkeyDonk

Originally posted by: desteffy

Originally posted by: HonkeyDonk

Originally posted by: desteffy
What do you mean by "solve" anyways.

That right there is a region. Just given a few inequalities like that it is possible that some are redundant, but not necessairly. Something like this can not generally be simplified down to saying x< some number and y > some number or whatever.

er...that's exactly what solve is supposed to mean.

When you are given two inequalities w/ 2 variables like in the OP's example, u can solve it so that you know X is greater than "a certain number", same with y.

You are correct about the "regions" and stuf, b/c that what it is when dealing w/ inequalities, but again, solving those inequalities means finding what the boundaries are i guess.

but then it wouldtn be called solving it. You may ask how to find an upper or lower bound for x or y, but saying that you can solve it like any other system:
ie

[system of equations] <=> x<number, y>number

is not correct. It just cant be done, just think of what the graph of it in the plane looks like.

Go read any math book involving inequalities (usually pre alegbra/algebra books) and they will say "solve this system of inequalities"

The world solve isn't strictly tied down to only finding a certain number "x" or "y", etc.

You can solve this set of inequalities by finding out what X is > or < (same goes for y). Simple as that.

No, it is NOT as simple as that.

They may want you to graph the solution set in the x-y plane, but you generally CAN NOT reduce systems like that to simple bounds on x and y.

 

[HonkeyDonk]

Originally posted by: DrPizza
desteffy is correct.

First of all, this isn't 6th grade math. It's typically taught in 8th or 9th grade. You were close, Honkeydonk. But, your solution indicates that you're in 8th grade (or perhaps 5th grade if it IS taught in 6th grade) because your solution is completely incorrect.

The solution isn't a particular value - the solution is an infinitely large set of points as can be simply depicted by a graph.

sorry i'm right, you just may not understand what I did in my solving.

I solved the set of inequalities so that it was in a form of what x and y are greater or less than.

If you plot those two inequalities, you will see that X and Y will not be less than a certain number (or possibly greater, I haven't graphed it yet).

Yes there are infinite solutions to what X and Y can be, I agree with you there, and again I agree that you can solve it by graphing it. However, solving it like the way I did will show what the values of X and Y have to be greater than (or less than).

You are too quick to judge my math skills. I graduated w/ a BSEE and have taken lots of math courses. I'm sure you couldn't care less though.

 

[HonkeyDonk]

Originally posted by: desteffy

Originally posted by: HonkeyDonk

Originally posted by: desteffy

Originally posted by: HonkeyDonk

Originally posted by: desteffy
What do you mean by "solve" anyways.

That right there is a region. Just given a few inequalities like that it is possible that some are redundant, but not necessairly. Something like this can not generally be simplified down to saying x< some number and y > some number or whatever.

er...that's exactly what solve is supposed to mean.

When you are given two inequalities w/ 2 variables like in the OP's example, u can solve it so that you know X is greater than "a certain number", same with y.

You are correct about the "regions" and stuf, b/c that what it is when dealing w/ inequalities, but again, solving those inequalities means finding what the boundaries are i guess.

but then it wouldtn be called solving it. You may ask how to find an upper or lower bound for x or y, but saying that you can solve it like any other system:
ie

[system of equations] <=> x<number, y>number

is not correct. It just cant be done, just think of what the graph of it in the plane looks like.

Go read any math book involving inequalities (usually pre alegbra/algebra books) and they will say "solve this system of inequalities"

The world solve isn't strictly tied down to only finding a certain number "x" or "y", etc.

You can solve this set of inequalities by finding out what X is > or < (same goes for y). Simple as that.

No, it is NOT as simple as that.

They may want you to graph the solution set in the x-y plane, but you generally CAN NOT reduce systems like that to simple bounds on x and y.

i agree, graphing is usually the best way to show these types of problems.

However, look at my solution and you will see that my answers are correct. Use Y > -16/6, so you can pick like 5 if you want. Then plug that into the first eqn and you will see that it is valid. Thus my solution works for all Y > -16/6.

 

[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.

 

[IEC]

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

 

[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.