Solve Inequalities Problem: x ∈ (-2,-1) ∪ (⅔, -½)

In summary, the notation "x ∈ (-2,-1) ∪ (⅔, -½)" indicates that the variable x belongs to the set of all real numbers between -2 and -1, not including -2 and -1, as well as the set of all real numbers between ⅔ and -½, not including ⅔ and -½. To graph this inequality, plot the two intervals on a number line and darken the sections in between. The solution set is the set of all real numbers between the two intervals, and to solve an inequality problem involving sets, identify the sets and their relationship and solve them separately, combining the solutions to determine the overall solution set. While there is no specific shortcut, familiarity
  • #1
erisedk
374
7

Homework Statement


Find the set of all ##x## for which ##\dfrac{2x}{2x^2 + 5x + 2} > \dfrac{1}{x + 1}##

Homework Equations

The Attempt at a Solution

I'm getting two different sets of answers with two different methods:Method 1-Wrong

##\dfrac{2x}{2x^2 + 5x + 2} > \dfrac{1}{x + 1}####\dfrac{2x^2 + 5x + 2}{2x} < x + 1####\dfrac{2x^2 + 5x + 2}{2x} - (x+1) < 0####\dfrac{2x^2 + 5x + 2 - 2x(x+1)}{2x} < 0####\dfrac{3x + 2}{2x} < 0#### x \in \left( \dfrac{-2}{3}, 0 \right)##
Method 2, the correct one

##\dfrac{2x}{2x^2 + 5x + 2} > \dfrac{1}{x + 1}####\dfrac{2x}{2x^2 + 5x + 2} - \dfrac{1}{x + 1} > 0####\dfrac{2x(x+1) - (2x^2 + 5x + 2)}{(2x^2 + 5x + 2)(x + 1)} > 0####\dfrac{3x + 2}{(2x + 1)(x + 1)(x + 2)} < 0 ####x \in (-2,-1) \cup \left(\dfrac{-2}{3} , \dfrac{-1}{2}\right)##
 
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  • #2
erisedk said:

Homework Statement


Find the set of all ##x## for which ##\dfrac{2x}{2x^2 + 5x + 2} > \dfrac{1}{x + 1}##

Homework Equations

The Attempt at a Solution

I'm getting two different sets of answers with two different methods:Method 1-Wrong

##\dfrac{2x}{2x^2 + 5x + 2} > \dfrac{1}{x + 1}####\dfrac{2x^2 + 5x + 2}{2x} < x + 1####\dfrac{2x^2 + 5x + 2}{2x} - (x+1) < 0####\dfrac{2x^2 + 5x + 2 - 2x(x+1)}{2x} < 0####\dfrac{3x + 2}{2x} < 0#### x \in \left( \dfrac{-2}{3}, 0 \right)##
Method 2, the correct one

##\dfrac{2x}{2x^2 + 5x + 2} > \dfrac{1}{x + 1}####\dfrac{2x}{2x^2 + 5x + 2} - \dfrac{1}{x + 1} > 0####\dfrac{2x(x+1) - (2x^2 + 5x + 2)}{(2x^2 + 5x + 2)(x + 1)} > 0####\dfrac{3x + 2}{(2x + 1)(x + 1)(x + 2)} < 0 ####x \in (-2,-1) \cup \left(\dfrac{-2}{3} , \dfrac{-1}{2}\right)##
In method 1, the first step is correct only if the left hand side and right hand side have the same sign.

If the left hand side is positive and the right hand side is negative, then which way does the inequality go after taking the reciprocal?
 
  • #3
Oh! Got it, thank you :)
 
  • #4
erisedk said:

Homework Statement


Find the set of all ##x## for which ##\dfrac{2x}{2x^2 + 5x + 2} > \dfrac{1}{x + 1}##

Homework Equations

The Attempt at a Solution

I'm getting two different sets of answers with two different methods:

Method 2, the correct one

##\dfrac{2x}{2x^2 + 5x + 2} > \dfrac{1}{x + 1}##
##\dfrac{2x}{2x^2 + 5x + 2} - \dfrac{1}{x + 1} > 0##
##\dfrac{2x(x+1) - (2x^2 + 5x + 2)}{(2x^2 + 5x + 2)(x + 1)} > 0##
##\dfrac{3x + 2}{(2x + 1)(x + 1)(x + 2)} < 0 ##

##x \in (-2,-1) \cup \left(\dfrac{-2}{3} , \dfrac{-1}{2}\right)##

How do you know you can go from ##\dfrac{2x}{2x^2 + 5x + 2} - \dfrac{1}{x + 1} > 0## to ##\dfrac{2x(x+1) - (2x^2 + 5x + 2)}{(2x^2 + 5x + 2)(x + 1)} > 0##? Certainly, if you multiply both sides of an inequality by a positive quantity, the inequality remains unchanged in direction. However, if you multiply both sides by a negative, the inequality is reversed.
 
  • #5
Ray Vickson said:
How do you know you can go from ##\dfrac{2x}{2x^2 + 5x + 2} - \dfrac{1}{x + 1} > 0## to ##\dfrac{2x(x+1) - (2x^2 + 5x + 2)}{(2x^2 + 5x + 2)(x + 1)} > 0##? Certainly, if you multiply both sides of an inequality by a positive quantity, the inequality remains unchanged in direction. However, if you multiply both sides by a negative, the inequality is reversed.
It looks more like OP combined rational expressions by using a common denominator, rather than by multiplying the whole expression by anything.
 
  • #6
SammyS said:
It looks more like OP combined rational expressions by using a common denominator, rather than by multiplying the whole expression by anything.

Agreed, but I would have preferred that the OP answer the question.
 

Related to Solve Inequalities Problem: x ∈ (-2,-1) ∪ (⅔, -½)

1. What does the notation "x ∈ (-2,-1) ∪ (⅔, -½)" mean?

The notation indicates that the variable x belongs to the set of all real numbers between -2 and -1, not including -2 and -1, as well as the set of all real numbers between ⅔ and -½, not including ⅔ and -½.

2. How do I graph x ∈ (-2,-1) ∪ (⅔, -½)?

To graph this inequality, first plot the two intervals on a number line. Then, darken the sections between -2 and -1 and between ⅔ and -½ to represent that x can take on any value within these intervals. The graph should resemble two darkened intervals with a gap in between them.

3. What is the solution set for x ∈ (-2,-1) ∪ (⅔, -½)?

The solution set is the set of all real numbers between -2 and -1, not including -2 and -1, as well as the set of all real numbers between ⅔ and -½, not including ⅔ and -½. In interval notation, the solution set can be written as (-2,-1) ∪ (⅔, -½).

4. How do I solve an inequality problem involving sets?

To solve an inequality problem involving sets, first identify the sets involved and the relationship between them (union, intersection, etc.). Then, solve the inequality for each set separately and combine the solutions to determine the overall solution set.

5. Is there a shortcut or easier way to solve an inequality problem involving sets?

In general, there is no specific shortcut or easier way to solve an inequality problem involving sets. However, it may be helpful to familiarize yourself with the properties and rules of set operations, as well as practice with various examples to become more proficient in solving these types of problems.

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