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

[erisedk]

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)$

 

[SammyS]

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?

 

[erisedk]

Oh! Got it, thank you :)

 

[Ray Vickson]

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.

 

[SammyS]

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.

 

[Ray Vickson]

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.