- #1

solakis1

- 425

- 0

Solve the following system of imequalities:

2x+3y-4>0

3x-4y+5>0

2x+3y-4>0

3x-4y+5>0

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- MHB
- Thread starter solakis1
- Start date

- #1

solakis1

- 425

- 0

Solve the following system of imequalities:

2x+3y-4>0

3x-4y+5>0

2x+3y-4>0

3x-4y+5>0

- #2

- #3

solakis1

- 425

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How about an algebraic solutionhttps://www.physicsforums.com/attachments/10382

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- #4

skeeter

- 1,104

- 1

How about an algebraic solution

you posted a challenge problem without knowing the solution beforehand?

for $x > \dfrac{1}{17}$ , $\dfrac{4-2x}{3} < y < \dfrac{3x+5}{4}$

- #5

kaliprasad

Gold Member

MHB

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secondly it is too simple to be in challenging section.

- #6

anemone

Gold Member

MHB

POTW Director

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I suppose you never realized you have been posted in the "Challenges" subforum.;) That is okay. No worries.

But I wish to tell you that in this subforum, if you posted a challenge problem here, that usually means you found a problem that is really intriguing and that you have solved it and wanted to share it with the community. In this case, your post is a challenge thread. Or, you found a problem (not a typical math problem) that intrigued you but you couldn't solve it, yet you are interested to know how other people might solve it, you could post that problem here and claimed it as Unsolved Challenge. I hope I have made myself clear.(Smile)

I welcome you to post again at the appropriate forum if you have any additional or future questions.

- #7

solakis1

- 425

- 0

(22/17)<y<infinity and (4/3)y-85/51<x<infinity

or

-infinity<y<(22/17) and -(3/2)y +68/34<x<infity

Later i will show in details how i got that solution.

Sorry for the delay but i had a lot of work to do

- #8

solakis1

- 425

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The solution of the system: 2x+3y-4=0, 3x-4y+5=0 are:

x=1/17 and y=22/17.

x=1/17 and y=22/17.

- #9

solakis1

- 425

- 0

(22/17)<y<infinity and (4/3)y-85/51<x<infinity

or

-infinity<y<(22/17) and -(3/2)y +68/34<x<infity

Later i will show in details how i got that solution.

Sorry for the delay but i had a lot of work to do

[sp]The solution of the system: 2x+3y-4=0, 3x-4y+5=0 are:

x=1/17 and y=22/17.

Now put:

x=1/17+n and y =22/17 +m ............(1)

and substituting those back to the original inequalities we get:

2m+3n>0 and 3m-4n>0

or

m>-(2/3)n and m>(4/3)n.........(2)

Now we have the following cases:

a)n is a positive No.Then from (2) we have:

m>(4/3)n and from this inequality and (1) we have:

22/17<y<infinity and (4/3)y-(85/51)<x<infinity.........(3)

b)for n negative ,then from (2) we have :

m>-(3/2)n,and from this inequality and (1) we have:

-infinity<y<22/17 and-(3/2)y+68/34<x<infinity..........(4)

Hence the solution of the inequalities are (3 ) ,(4)[/sp]

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