Question

Complete solution set of inequality
(x+2)(x+3)/(x-2)(x-3)< or = 1is​

Answer 1

Step-by-step explanation:

We have,

$\frac{(x + 2)(x + 3)}{(x - 2)(x - 3)} \leqslant 1$

$\implies \frac{(x + 2)(x + 3)}{(x - 2)(x - 3)} - 1 \leqslant 0$

$\implies \frac{(x^{2} + 5x + 6) }{( {x}^{2} - 5x + 6) } - 1 \leqslant 0$

$\implies \frac{(x^{2} + 5x + 6) - ( {x}^{2} - 5x + 6)}{( {x}^{2} - 5x + 6) } \leqslant 0$

$\implies \frac{(x^{2} + 5x + 6 - {x}^{2} + 5x - 6)}{( {x}^{2} - 5x + 6) } \leqslant 0$

$\implies \frac{10x}{ {x}^{2} - 5x + 6 } \leqslant 0$

$\implies \frac{10x}{ (x - 3)(x - 2)} \leqslant 0$

$x \in( - \infty ,0 ] \cup(2,3)$

Answer 2

Answer:

The solution of given inequality is

$x\epsilon(-\infty,0]U(2,3)$

Step-by-step explanation:

The given inequality is

$\frac{(x + 2)(x + 3)}{(x - 2)(x - 3)} \leqslant 1$

Subtracting 1 on both sides,we get

$\frac{(x + 2)(x + 3)}{(x - 2)(x - 3)} - 1\leqslant 1 - 1 \\ = > \frac{ {x}^{2} + 5x + 6 }{ {x}^{2} - 5x + 6 } - 1 \leqslant 0 \\ = > \frac{10x}{ {x}^{2} - 5x + 6 } \leqslant 0$

After factorising the denominator,we can rewrite the above inequality as

$\frac{10x}{(x - 2)(x - 3)} \leqslant 0$

To solve this we shall use an concept as shown

$\frac{a}{b} \leqslant 0 \\ = > a \leqslant 0 \: b > 0 \: or \\ a \geqslant 0 \: b < 0$

If we consider first condition,we get

$x \leqslant 0 \\ (x - 2)(x - 3) > 0 = > x < 2 \: and \: x > 3$

If we consider second condition,we get

$x \geqslant 0 \\ (x - 2)(x - 3) < 0 = > 2 < x < 3$

Hence after obtaining intersection of the above inequalities we get the solution of inequality as

$x\epsilon(-\infty,0]U(2,3)$

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