Question

Solve the system of inequations
(x+3)/(x-2) <=2,(2x+5)/(x+7)>=3​

Answer 1

$\large\underline{\sf{Solution-}}$

Given system of in-equations are

$\rm :\longmapsto\:\dfrac{x + 3}{x - 2} \leqslant 2$

and

$\rm :\longmapsto\:\dfrac{2x + 5}{x + 7} \geqslant 3$

Consider,

$\rm :\longmapsto\:\dfrac{x + 3}{x - 2} \leqslant 2$

$\rm :\longmapsto\:\dfrac{x + 3}{x - 2} - 2 \leqslant 0 \: \: and \: x \: \ne \: 2$

$\rm :\longmapsto\:\dfrac{x + 3 -2 x + 4}{x - 2} \leqslant 0 \: \: and \: x \: \ne \: 2$

$\rm :\longmapsto\:\dfrac{ - x + 7}{x - 2} \leqslant 0 \: \: and \: x \: \ne \: 2$

$\rm :\longmapsto\:\dfrac{x - 7}{x - 2} \geqslant 0 \: \: and \: x \: \ne \: 2$

We know,

If a < b

$\boxed{ \bf{ \: \frac{x - a}{x - b} \geqslant 0 \: \bf\implies \:x \leqslant a \:or \: x > b}}$

So, using this,

$\bf\implies \:x < 2 \: \: or \: \: x \geqslant 7 - - - (1)$

Consider,

$\rm :\longmapsto\:\dfrac{2x + 5}{x + 7} \geqslant 3$

$\rm :\longmapsto\:\dfrac{2x + 5}{x + 7} - 3\geqslant 0 \: \: \: and \: x \: \ne \: - 7$

$\rm :\longmapsto\:\dfrac{2x + 5 - 3x - 21}{x + 7} \geqslant 0 \: \: \: and \: x \: \ne \: - 7$

$\rm :\longmapsto\:\dfrac{- x - 16}{x + 7} \geqslant 0 \: \: \: and \: x \: \ne \: - 7$

$\rm :\longmapsto\:\dfrac{-( x + 16)}{x + 7} \geqslant 0 \: \: \: and \: x \: \ne \: - 7$

$\rm :\longmapsto\:\dfrac{x + 16}{x + 7} \leqslant 0 \: \: \: and \: x \: \ne \: - 7$

We know,

If a < b

$\boxed{ \bf{ \: \frac{x - a}{x - b} \leqslant 0 \: \bf\implies \:a \leqslant x < b}}$

So, using this,

$\bf\implies \ - 16 \leqslant x < - 7 - - - (2)$

From equation (1) and (2), we concluded that

$\bf\implies \:x \: \in \: [- 16 , - 7)$

Answer 2

Answer:

$\large\underline{\sf{Solution-}}Solution−</p><p></p><p>Given system of in-equations are</p><p></p><p>\rm :\longmapsto\:\dfrac{x + 3}{x - 2} \leqslant 2:⟼x−2x+3⩽2</p><p></p><p>and</p><p></p><p>\rm :\longmapsto\:\dfrac{2x + 5}{x + 7} \geqslant 3:⟼x+72x+5⩾3</p><p></p><p>Consider,</p><p></p><p>\rm :\longmapsto\:\dfrac{x + 3}{x - 2} \leqslant 2:⟼x−2x+3⩽2</p><p></p><p>\rm :\longmapsto\:\dfrac{x + 3}{x - 2} - 2 \leqslant 0 \: \: and \: x \: \ne \: 2:⟼x−2x+3−2⩽0andx=2</p><p></p><p>\rm :\longmapsto\:\dfrac{x + 3 -2 x + 4}{x - 2} \leqslant 0 \: \: and \: x \: \ne \: 2:⟼x−2x+3−2x+4⩽0andx=2</p><p></p><p>\rm :\longmapsto\:\dfrac{ - x + 7}{x - 2} \leqslant 0 \: \: and \: x \: \ne \: 2:⟼x−2−x+7⩽0andx=2</p><p></p><p>\rm :\longmapsto\:\dfrac{x - 7}{x - 2} \geqslant 0 \: \: and \: x \: \ne \: 2:⟼x−2x−7⩾0andx=2</p><p></p><p>We know,</p><p></p><p>If a < b</p><p></p><p>\boxed{ \bf{ \: \frac{x - a}{x - b} \geqslant 0 \: \bf\implies \:x \leqslant a \:or \: x > b}}x−bx−a⩾0⟹x⩽aorx>b</p><p></p><p>So, using this,</p><p></p><p>\bf\implies \:x < 2 \: \: or \: \: x \geqslant 7 - - - (1)⟹x<2orx⩾7−−−(1)</p><p></p><p>Consider,</p><p></p><p>\rm :\longmapsto\:\dfrac{2x + 5}{x + 7} \geqslant 3:⟼x+72x+5⩾3</p><p></p><p>\rm :\longmapsto\:\dfrac{2x + 5}{x + 7} - 3\geqslant 0 \: \: \: and \: x \: \ne \: - 7:⟼x+72x+5−3⩾0andx=−7</p><p></p><p>\rm :\longmapsto\:\dfrac{2x + 5 - 3x - 21}{x + 7} \geqslant 0 \: \: \: and \: x \: \ne \: - 7:⟼x+72x+5−3x−21⩾0andx=−7</p><p></p><p>\rm :\longmapsto\:\dfrac{- x - 16}{x + 7} \geqslant 0 \: \: \: and \: x \: \ne \: - 7:⟼x+7−x−16⩾0andx=−7</p><p></p><p>\rm :\longmapsto\:\dfrac{-( x + 16)}{x + 7} \geqslant 0 \: \: \: and \: x \: \ne \: - 7:⟼x+7−(x+16)⩾0andx=−7</p><p></p><p>\rm :\longmapsto\:\dfrac{x + 16}{x + 7} \leqslant 0 \: \: \: and \: x \: \ne \: - 7:⟼x+7x+16⩽0andx=−7</p><p></p><p>We know,</p><p></p><p>If a < b</p><p></p><p>\boxed{ \bf{ \: \frac{x - a}{x - b} \leqslant 0 \: \bf\implies \:a \leqslant x < b}}x−bx−a⩽0⟹a⩽x<b</p><p></p><p>So, using this,</p><p></p><p>\bf\implies \ - 16 \leqslant x < - 7 - - - (2)⟹ −16⩽x<−7−−−(2)</p><p></p><p>From equation (1) and (2), we concluded that</p><p></p><p>\bf\implies \:x \: \in \: [- 16 , - 7)⟹x∈[−16,−7)</p><p>$