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Solving inequalities with rational expressions.

  • Thread starter viet1919
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  • #1
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Homework Statement


Identify the solution set of the inequality.

Homework Equations


5x + 1 / x- 1 ≥ 7


The Attempt at a Solution


I multiplied both sides by x - 1 which gave me 5x+1 ≥ 7x - 7
Then I combined like terms together which gave me 8 ≥ 2x
I divided both sides by 2 which gave me 4 ≥ x.
Now that's not the answer. The answer is 1 < x ≤ 4.
Why is that?
 

Answers and Replies

  • #2
Simon Bridge
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Substitute x=1 into the first relation and see what happens?
What happens to it when x < 1?
 
  • #3
ehild
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Homework Statement


Identify the solution set of the inequality.

Homework Equations


5x + 1 / x- 1 ≥ 7


The Attempt at a Solution


I multiplied both sides by x - 1 which gave me 5x+1 ≥ 7x - 7
Then I combined like terms together which gave me 8 ≥ 2x
I divided both sides by 2 which gave me 4 ≥ x.
Now that's not the answer. The answer is 1 < x ≤ 4.
Why is that?
Hi Viet, welcome to PF.

Did you mean [itex]5x+\frac{1}{x}-1\ge7[/itex] or [itex]\frac{5x+1}{x-1}\ge7[/itex] ? Do not forget the parentheses!!!
Multiplying both sides of an inequality with a negative quantity will flip the inequality over. So you keep the inequality when multiplying by x-1 if x>1, and then you get x ≤4.

Assuming x<1 you get contradiction.

ehild
 
  • #4
Ray Vickson
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Homework Statement


Identify the solution set of the inequality.

Homework Equations


5x + 1 / x- 1 ≥ 7


The Attempt at a Solution


I multiplied both sides by x - 1 which gave me 5x+1 ≥ 7x - 7
Then I combined like terms together which gave me 8 ≥ 2x
I divided both sides by 2 which gave me 4 ≥ x.
Now that's not the answer. The answer is 1 < x ≤ 4.
Why is that?
Your inequality---as written--- is
[tex] f(x) = 5x + \frac{1}{x} -1 \geq 7 [/tex]
The 'boundary' points are where f(x) = 7 exactly, so are given by the roots of the quadratic equation ##5x^2+1 = 8x,## or
[tex]x = \frac{4}{5} \pm \frac{\sqrt{11}}{5} \doteq 0.133675, \:1.463325 [/tex]
Note that f(x) < 0 for x < 0, so we need x ≥ 0. For ##x \to 0+## or ##x \to +\infty## we have ##f(x) \to + \infty > 7,## so the solution is ##0 < x \leq 0.144675## and ##x \geq 1.463325##.

All this is under the assumption that what you wrote is what you meant when read using standard rules. If you had meant (5x + 1)/(x-1) ≥ 7 you would have used parentheses.
 
  • #5
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[itex]\frac{5x + 1}{x - 1}[/itex] ≥ 7
That is what I meant. Sorry. My first time using the forums. I didn't know what I was doing. Now I am still confused on how to solve this. As I mentioned before, in my method of solving this I ended up with a one inequality solution: x ≥ 4
Why is that that's not the answer? The answer is 1 < x ≤ 4. Where did the 1 come from?
 
  • #6
Simon Bridge
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[itex]\frac{5x + 1}{x - 1}[/itex] ≥ 7
That is what I meant. Sorry. My first time using the forums. I didn't know what I was doing. Now I am still confused on how to solve this. As I mentioned before, in my method of solving this I ended up with a one inequality solution: x ≥ 4
Why is that that's not the answer? The answer is 1 < x ≤ 4. Where did the 1 come from?
Please reread post #2 - try the suggestion and you'll see where the 1 came from.
Please reread post #3 to see the origin of your mistake.

Your mistake was to fail to take account of the situation where (x-1)≤0.
 
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  • #7
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I THINK understand now. Thank you.
So...
X is not less than 1 because they are the same value making X greater than 1. Is that right?
 
  • #8
ehild
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When you multiply an inequality with a negative quantity, it flips over.

See an example:

-2<1 is true.

Multiply both sides by -1: you get 2 < -1 which is false. You have to change "<" to ">".

2>-1.

If x-1<0 and you multiply with it, the inequality becomes 5x-1≤7(x-1) --> 6≤2x, x>3 , but you assumed that x was less than 1, so you arrived at contradiction. x can not be less than 1 and greater than 3 at the same time.

X also cannot be 1, as it would mean zero in the denominator. So x>1.

ehild
 
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  • #9
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I completely understand now. Thank you very much and sorry for such a late reply. What you did was work with each equation individually. Sorry fro not understanding so fast. My brain takes in things differently than most people. When using math terms, I'm not that well at interpreting it unless it's shown. Thank you.
 
  • #10
ehild
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I think most people understand Maths through simple examples. When in doubt doing something, try it on simple numerical examples. It helps.


ehild
 

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