madz_08
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I'm doing a question with inequalities which i initially thought was very simple...

"solve the inequality ((2x-5)/(x+1))<1"

I solved it as if < is a = and eventually got x<6 (which is correct) but there were additional workings to find -1<x which i dont get where that has come from??

In the workings they made the inequality into a quadratic inequality, which i get how they get there, but couldn't you just multiply by (x+1) and get a simple linear inequality?

help appreciated thanks )
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mqb2766
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(Original post by madz_08)
I'm doing a question with inequalities which i initially thought was very simple...

"solve the inequality ((2x-5)/(x+1))<1"

I solved it as if < is a = and eventually got x<6 (which is correct) but there were additional workings to find -1<x which i dont get where that has come from??

In the workings they made the inequality into a quadratic inequality, which i get how they get there, but couldn't you just multiply by (x+1) and get a simple linear inequality?

help appreciated thanks )
1) What did you assume when you multiplied through by (x+1)?
2) You can, but dont have to, use a quadratic, but you have to be careful with point 1)
Last edited by mqb2766; 2 months ago
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madz_08
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(Original post by mqb2766)
1) What did you assume when you multiplied through by (x+1)?
2) You can, but dont have to, use a quadratic, but you have to be careful with point 1)
1) not too sure...? Whenever i've solved inequalities ive always just thought of it as a normal equation and then tackle the <0 or >0 at the end (i.e drawing a graph and putting the inequalities in last)

still struggling a bit with the logic behind the -1 tbh
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mqb2766
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(Original post by madz_08)
1) not too sure...? Whenever i've solved inequalities ive always just thought of it as a normal equation and then tackle the <0 or >0 at the end (i.e drawing a graph and putting the inequalities in last)

still struggling a bit with the logic behind the -1 tbh
For the logic behind the -1 part, you could recognize the problem is roughly
1/x < 1
This has solutions
x>1
x<0
Just sketch it.

What happens when you multiply an inequality by a negative number?
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madz_08
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(Original post by mqb2766)
For the logic behind the -1 part, you could recognize the problem is roughly
1/x < 1
This has solutions
x>1
x<0
Just sketch it.

What happens when you multiply an inequality by a negative number?
You have to flip the inequality and hence x>-1 im assuming?
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mqb2766
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(Original post by madz_08)
You have to flip the inequality and hence x>-1 im assuming?
Don't assume. Youve found one solution by when x+1>0. What happens when x+1<0? Work it through.
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madz_08
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(Original post by mqb2766)
Don't assume. Youve found one solution by when x+1>0. What happens when x+1<0? Work it through.
Ok, I've reworked the question a few times now and getting a hold of it thanks very much for the help!!
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mqb2766
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(Original post by madz_08)
Ok, I've reworked the question a few times now and getting a hold of it thanks very much for the help!!
Good. Its always worthi sticking some values in, both positive and negative, and evaluate the original function(s) in the inequality. Which ones satisfy the inequality.
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