The process of solving rational inequalities is close to the method of solving polynomial inequality.

In solving the rational inequality ,special attention needs to be given to the denominator.

Never multiply both sides by denominator

Always bring all the terms to one side and make another side zero.

Do not ignore the zeros of denominator.

If the inequality sign has = included like $\leq$ or $\geq$, then check the intervals carefully. The zeros of the denominators will always be excluded from the solution as making denominator zero makes the expression undefined.

$\frac{2x-1}{x+1}$ $> 0$

The common practice is to multiply both sides by the denominator $x + 1$.

So,

$(2x - 1) > 0$

Solving for $x$, we get

$2x - 1 > 0$

$x >$ $\frac{1}{2}$

Let us check the inequality for region

$x$ = $3$, $\frac{2x-1}{x+1}$ = $\frac{6-1}{3+1}$ $> 0$ True

Check if $x$ = $-2$.

$\frac{2x + 1}{x + 1}$ = $\frac{-4 + 1}{-2 + 1}$ = $\frac{-3}{-1}$ = $2 > 0$

Thus as per the solution obtained the solution has to be $x > 2$ but the value $x$ = $-2$ does not belong to the interval $(2, \infty)$ which implies that the method of solving rational inequality is incorrect .

The main concept forgotten here is that if an inequality is multiplied of divided by the negative number, its inequality sign reverses. Since $x + 1$ has a variable in it which can take any value, $x + 1$ can be negative as well.

Considering $x + 1$ as negative , the inequality $\frac{2x-1}{x+1}$ $> 0$ becomes $\frac{2x - 1}{x + 1}$ $\times\ (x + 1) < 0\ \times\ (x + 1)$

$2x - 1 < 0$

$x <$ $\frac{1}{2}$

$x + 1$ = $0$ will make the expression undefined so $x$ = $-1$ becomes restricted value.

Proper Step by Step method to solve Rational Inequality:

Subtract $\frac{2}{x}$ from both sides.

$\frac{x-3}{x+3}$ -3 - $\frac{2}{x}$ < $\frac{2}{x}$ - $\frac{2}{x}$

$\frac{x-3}{x+3}$ - 3 - $\frac{2}{x}$ < 0

Simplify the left side.

The LCD for $x$ and $x + 3$ will be $x(x + 3)$.

$\frac{x-3}{x+3}$ $\frac{x}{x}$ - 3 $\frac{x \times (x+3)}{x \times (x+3)}$ - $\frac{2}{x}$ $\times$ $\frac{x+3}{x+3}$ < 0

$\frac{x^2-3x}{x \times (x+3)}$ - $\frac{3x^2-9x}{x \times (x+3)}$ - $\frac{2x-6}{x \times (x+3)}$ < 0

$\frac{x^2-3x-3x^2+9x-2x+6}{x \times (x+3)}$ < 0

$\frac{-2x^2+4x+6}{x \times (x+3)}$ < 0

Find zeros of numerator and the denominator separately.

$-2x^2 + 4x + 6$ = $0$ and $x(x + 3)$ = $0$

To have $\frac{-2x^2+4x+6}{x \times (x+3)}$ <0 either of the numerator or the denominator should be negative.

Its better to use the number line sign chart to check for the inequality interval.

$-2x^2 + 4x + 6$ = $0$ $x(x + 3)$ = $0$

$x^2 - 2 - 3$ = $0$ $x$ = $0$ or $x + 3$ = $0$

$(x - 3)(x + 1)$ = $0$ $x$ = $0$ or $x$ = $-3$

$x - 3$ = $0$ or $x + 1$ = $0$

So, finally we get

$x$ = $3,\ x$ = $-1,\ x$ = $0$ and $x$ = $-3$.

Mark all these points on the number line and then check the sign of the expression in the given region.

Check $(-\infty,\ -3),\ x$ = $-4$

$\frac{-2(x-3)(x+1)}{x \times (x+3)}$ = $\frac{-2(-4-3)(-4+1)}{-4 \times (-4+3)}$ = $\frac{-ve}{+ve}$

So, the expression is negative.

Next region is $(-3,\ -1)$ Take a value in this interval,say $x$ = $-2$

The expression will be $-2 \times -5 \times$ $-\frac{1}{-2}$ $\times 1$: $\frac{-ve}{+ve}$: $+ve$ so $> 0$

Similarly check for all the intervals.

Since we need $\frac{-2x^2+4x+6}{x \times (x+3)}$ < 0

The intervals that have negative sign will be the solution.

The solution will be $(-\infty,\ -3) (-1,\ 0)$ and $(3,\ 0)$

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