Solving inequalities with rational expressions.

In summary, an inequality can only have one solution if the inequality is reversed (x-1>0). 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.
  • #1
viet1919
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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?
 
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  • #2
Substitute x=1 into the first relation and see what happens?
What happens to it when x < 1?
 
  • #3
viet1919 said:

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
viet1919 said:

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
[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
viet1919 said:
[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
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
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
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
I think most people understand Maths through simple examples. When in doubt doing something, try it on simple numerical examples. It helps. ehild
 

What is a rational expression?

A rational expression is an algebraic expression that is written as a ratio of two polynomials. It can also be thought of as a fraction with variables in the numerator and/or denominator.

How do you solve inequalities with rational expressions?

To solve an inequality with rational expressions, you first need to find the values that make the denominator(s) equal to zero. These values are called the excluded values. Then, solve the resulting inequality by using the same steps as solving an algebraic inequality.

What are excluded values?

Excluded values are values that make the denominator(s) of a rational expression equal to zero. These values must be excluded from the solution set because they would result in division by zero, which is undefined.

Are there any restrictions when solving inequalities with rational expressions?

Yes, there are a few restrictions when solving inequalities with rational expressions. The first is that you cannot multiply or divide both sides of the inequality by a negative number, as this would change the direction of the inequality. Additionally, you must remember to exclude any values that make the denominator(s) zero.

Can rational inequalities have more than one solution?

Yes, rational inequalities can have multiple solutions. This is because the solution set may contain values that make the inequality true, but do not make the denominator(s) equal to zero. Additionally, the solution set may also contain excluded values that must be excluded from the final solution.

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