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# Solving inequality watch

1. Solve for x :

How would you go about solving this? Multiply by {x-1}^2 and {x+1}^2 ?
2. Christ, I am in awe of anyone who attempts maths during the holidays! I'm sorry but my brain just won't work between July and September
3. Multiply by (x - 1)(x + 1), multiply out the brackets, things cancel, stick all the x's on one side.
Multiply by (x - 1)(x + 1), multiply out the brackets, things cancel, stick all the x's on one side.
x^2 - 1 is sometimes negative, which would reverse the sign of the inequality
5. (Original post by around)
x^2 - 1 is sometimes negative, which would reverse the sign of the inequality
Oh yeah, how embarrassing. So if I remember right then you do it how I said, but for the case of x>1 and then again for x<1.
6. Well ive subtract the RHS on the LHS then put them together and finally got :

so if it i was = 0 id have x=1, x=-1 and x=-1/7

so for > 0 what will the interval be?
7. Sugar , is this GCSE ?
8. IB Higher Level, and i dont think we can solve using a GDC

9. Cross multiply, collect terms, etc.
10. (Original post by D-Day)

Cross multiply, collect terms, etc.

But we dont know if x+1 or x-1 is negative so would change the sign.
11. Ive solved a different way and got

but still unsure what the interval for x will be/
12. (Original post by Coded)
But we dont know if x+1 or x-1 is negative so would change the sign.
Changing the sign only happens when dividing by a negative constant. Here you are multiplying by an expression.
13. I got
14. (Original post by Coded)
Ive solved a different way and got

but still unsure what the interval for x will be/
x^2 - 1 is positive for x > 1 or x < -1, and negative for -1 < x < 1. 7x + 1 is positive for x > -1/7, negative for x < -1/7. So in which intervals will their product be positive?

(Original post by D-Day)
I got
Disagree. Try x = 0.
15. ^ No, at D-Day (x=0 doesn't work!). The answer is, I believe, x>1 or -1<x<-1/7

Just solve cases for positive and negative x!

EDIT: Damn you GE! Always quicker!
16. (Original post by generalebriety)
x^2 - 1 is positive for x > 1 or x < -1, and negative for -1 < x < 1. 7x + 1 is positive for x > -1/7, negative for x < -1/7. So in which intervals will their product be positive?
im liking that, that you.

so x > 1 or -1 < x < -1/7
17. (Original post by generalebriety)
Disagree. Try x = 0.
18. You can't directly re-arrange it as an inequality with x as the subject because x-1 and x+1 can be either negative or positive

You need to solve it as an equality to find the critical x values, then either draw a graph with both curves on it to find the intervals. (you could also use a table of values I think, but I've always used graphs for things like this so I don't forget to take asymptotes into account)
19. Or you can do what OP originally suggested (as none of those expressions are ever negative)

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