# (x+1)/(x-2)<(x-1)/(x+2)?

(x + 1)/(x - 2) < (x - 1)/(x + 2)

Cross multiplication.

(x + 1)(x + 2)<(x - 1)(x - 2).

FOIL method: the product of two binomials is the sum of the products of the First terms, the Outer terms, the Inner terms and the Last terms.

x2 + 2x + x + 2 < x2 - 2x - x + 2

x2 + 3x + 2 < x2 - 3x + 2

Subtract x2 & 2 from each side.

3x < - 3x

6x < 0

Divide each side by 6.

x < 0

Graph the solution set on a number line.

(x + 1)/(x − 2) < (x − 1)/(x + 2)
(x + 1)/(x − 2) − (x − 1)/(x + 2) < 0
((x + 1)(x + 2) − (x − 2)(x − 1))/((x − 2)(x + 2)) < 0
(x² + 3x + 2 − x² + 3x − 2)/((x − 2)(x + 2)) < 0
(6x)/((x − 2)(x + 2)) < 0

Roots of the numerator:
6x = 0 ⇒ x = 0. For x > 0, the numerator is positive. For x < 0, the numerator is negative. (crescent function)

Roots of the denominator (remember, it CAN'T be zero):
(x − 2)(x + 2) ≠ 0 ⇒ x ≠ 2, x ≠ −2. For |x| > 2, we have that the denominator is positive. Reciprocally, for |x| < 2, the denominator is negative.

So, our goal is to have both the numerator and the denominator with different signs (so it assumes a negative value, that is less than 0).

x > 2: +/+ = +, not what we want.
2 > x > 0: +/− = −
−2 < x < 0: −/− = +, not what we want.
x < −2: −/+ = −

So, the solution is, in set notation:
S = {x ∈ ℝ | x < −2 or 0 < x < 2}

Or, in interval notation:
S = (∞⁻, −2) U (0, 2)

The rational inequality is .

When solving a rational inequality, begin by writing the inequality in general form with the rational expression

on the left and zero on the right.

Now, the rational inequality is .

Step - 1 :

State the exclude values, those are the values for which denominator is zero.

The exclude value of the inequality is 2 and - 2.

Step - 2 :

Solve the related equation

6x = 0

x = 0.

Solution of related equation x = 0.

Step - 3 :

Draw the vertical lines at the exclude values and at the solution to separate the number line into intervals.

Contd.......

Step - 4 :

Now test  sample values in each interval to determine whether values in the interval satisify the inequality.

 Test interval x - value Inequality Conclusion (- ∞, -2) x = - 3 True (- 2, 0) x = - 1 False (0, 2) x = 1 True (2, ∞) x = 3 False

Note that the original inequality contains a “ < ” symbol, We exlude it into set of solutions at x = - 2

Above statement is true.

x < - 2 is a solution of inequality.

The above conclude that the inequality is satisfied for all x - values in (- ∞, -2) and (0, 2).

This implies that the solution  of  the  inequality is  the  interval (- ∞, - 2) and (0, 2) . as shown in Figure below. Note that the original inequality contains a “ < ” symbol. This means that the solution set does not contain the endpoints of the test interval is (- ∞, - 2) .

Solution of the inequality is { x | x < - 2 or 0 < x < 2 }.

The interval notation form of inequality is (- ∞, - 2) U (0, 2).