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November 08, 2025, 03:58:29 am

Author Topic: Linear Graphs Question  (Read 1552 times)  Share 

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inneedoftherapy

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Linear Graphs Question
« on: October 16, 2013, 10:17:01 pm »
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Hey guys,
Is this right?

There are at least three y for every two x
2y ≥ 3x

Because in someone else's notes it said,

For every eight x, there are at least ten y
8x ≥ 10y

which seems to contradict it? Confused. Which one is right?
Also, what's a good way of figuring out how to interpret it. The wording seems to get me a fair bit! :S

thanks!

ellie7g

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Re: Linear Graphs Question
« Reply #1 on: October 19, 2013, 02:50:39 pm »
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I have the exact same problems!

plato

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Re: Linear Graphs Question
« Reply #2 on: October 19, 2013, 06:18:45 pm »
+1
Hey guys,
Is this right?

There are at least three y for every two x
2y ≥ 3x

Because in someone else's notes it said,

For every eight x, there are at least ten y
8x ≥ 10y

which seems to contradict it? Confused. Which one is right?
Also, what's a good way of figuring out how to interpret it. The wording seems to get me a fair bit! :S

thanks!

The wording of the question is confusing since it seems to treat x and y as physical, individual items as you would if the question had said "there are at least three apples for every two oranges".
If this is meant to relate to a linear programming situation, then x and y must be defined as variables which have value rather than a physical presence.

For example, if you think of x as "the number of apples" and y as the "number of oranges", then the question can refer to numbers such as in "there are at least 3 oranges for every two apples".

To find and write an inequation for this constraint, start of with some simple numbers that fit the requirement and from which you could work out the rule you want.
eg 1.  if x =2, then
eg 2.  if x = 6, then
etc

In each case,    or   
This is the same as

#J.Procrastinator

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Re: Linear Graphs Question
« Reply #3 on: October 19, 2013, 07:17:52 pm »
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Oh right, to check you could also sub 2 into x to get:

2y ≥ 3x
2y ≥ 3(2)
y≥ 3

So you're right
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