Solve Rational Inequality: Find Integer Roots [-2, 3]

[mafagafo]

Homework Statement

Find all integer roots that satisfy (3x + 1)/(x - 4) < 1.

The Attempt at a Solution

I would do this:

Make it an equation and find x such that (3x + 1)/(x - 4) = 1.

3x + 1 = x - 4
2x = -5
x = -5/2

Then check if the inequality is valid for values smaller than x and for values bigger than x.

But this approach is not good enough as I would get [-2, +∞) {integers} as my answer.

Any help would be really appreciated.

I think that the answer is [-2, 3] {integers}. But could only get this with a plot.

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What should I also do so that my method is valid for "rational" inequalities?

 

[LCKurtz]

Homework Statement

Find all integer roots that satisfy (3x + 1)/(x - 4) < 1.

The Attempt at a Solution

I would do this:

Make it an equation and find x such that (3x + 1)/(x - 4) = 1.

3x + 1 = x - 4
2x = -5
x = -5/2

Then check if the inequality is valid for values smaller than x and for values bigger than x.

But this approach is not good enough as I would get [-2, +∞) {integers} as my answer.

Any help would be really appreciated.

I think that the answer is [-2, 3] {integers}. But could only get this with a plot.

---

What should I also do so that my method is valid for "rational" inequalities?

Consider the two cases where $x<4$ and $x>4$ and work the inequalities separately.

 

[pasmith]

Homework Statement

Find all integer roots that satisfy (3x + 1)/(x - 4) < 1.

The Attempt at a Solution

I would do this:

Make it an equation and find x such that (3x + 1)/(x - 4) = 1.

3x + 1 = x - 4
2x = -5
x = -5/2

Then check if the inequality is valid for values smaller than x and for values bigger than x.

But this approach is not good enough as I would get [-2, +∞) {integers} as my answer.

<br /> \frac{3x + 1}{x - 4} = \frac{3(x-4) + 3(4) + 1}{x - 4} = 3 + \frac{13}{x - 4}.<br /> Thus if (3x + 1)/(x-4) &lt; 1 then 13/(x - 4) &lt; - 2. Clearly that can't be the case if x &gt; 4 (because then 13/(x - 4) &gt; 0 &gt; -2) so we must have x &lt; 4. Is there a lower bound?

 

[ehild]

(3x + 1)/(x - 4) < 1 can be written in the form
\frac{(3x+1)-(x-4)}{x-4}&lt;0

Simplified: \frac{2x+5}{x-4}&lt;0

When is the fraction negative?

ehild

 

[mafagafo]

---
(3x + 1)/(x - 4) = 1
3x + 1 = x - 4
2x = -5

x = -5/2

----
(3x + 1)/(x - 4) = 1
(3(x - 4) + 12 + 1) / (x - 4) = 1
3 + 13/(x - 4) = 1
13 / (x - 4) = -2

x = 4

----
Then I work with those?
(3x + 1)/(x - 4) < 1

- 8/3 >> false
- 5/2 >> false
- 7/3 >> true
    4 >> impossible
    5 >> false

So the valid integers are {-2, -1, 0, 1, 2, 3}?

 

[HallsofIvy]

Homework Statement

Find all integer roots that satisfy (3x + 1)/(x - 4) < 1.

The Attempt at a Solution

I would do this:

Make it an equation and find x such that (3x + 1)/(x - 4) = 1.

3x + 1 = x - 4
2x = -5
x = -5/2

Then check if the inequality is valid for values smaller than x and for values bigger than x.

An inequality can change direction where the two sides are equal or where the functions are discontinuous. Here, the first occurs where x= -5/2 and the second where x= 4. There are three intervals to be considered: x< -5/2, -5/2< x< 4, and x> 4.
x= -3< -5/2 and (3(-3)+ 1)/(-3- 4)= (-9+ 1)/(-7)= -8/-7 is greater than 1 so NO x< -5/2 satisfies the inequality. x= 0 is between -5/2 and 4. (3(0)+ 1)(0- 4)= -1/4 is less than 1. Every x between -5/2 and 4 satisfy the inequality. x= 5 is larger than 4 and (3(5)+ 1)/(5- 4)= 15/1 is larger than 1. The integer solutions are -2, -1, 0, 1, 2, and 3.

But this approach is not good enough as I would get [-2, +∞) {integers} as my answer.

Any help would be really appreciated.

I think that the answer is [-2, 3] {integers}. But could only get this with a plot.

---

What should I also do so that my method is valid for "rational" inequalities?

 

[mafagafo]

(3x + 1)/(x - 4) < 1 can be written in the form
\frac{(3x+1)-(x-4)}{x-4}&lt;0

Simplified: \frac{2x+5}{x-4}&lt;0

When is the fraction negative?

ehild

When {2x+5} &lt; 0 and {x-4} &gt; 0 or when {2x+5} &gt; 0 and {x-4} &lt; 0.

If {2x+5} &lt; 0, then 2x&lt;-5 and x&lt;-\frac{5}{2}.
and if {x-4} &gt; 0, then x &gt; 4. Thus, this is impossible.

If {2x+5} &gt; 0, then 2x&gt;-5 and x&gt;-\frac{5}{2}.
and if {x-4} &lt; 0, then x &lt; 4. Thus, S=\left\{x\in Z|-5 /2 &lt; x &lt; 4\right\}=\left\{x\in Z|-2 \le x \le 3\right\}

 

[mafagafo]

Big thanks to all of you, with special mention to HallsOfIvy for answering my question.

Q.: "What should I also do so that my method is valid for "rational" inequalities?"
A.: An inequality can change direction where the two sides are equal or where the functions are discontinuous.