Question

Solve for x, 2x +3/x+3=3x+2/x+1​

Answer 1

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

$\rm :\longmapsto\:\dfrac{2x + 3}{x + 3} = \dfrac{3x + 2}{x + 1}$

$\rm :\longmapsto\:(2x + 3)(x + 1) = (3x + 2)(x + 3)$

$\rm :\longmapsto\:2x(x + 1) + 3(x + 1)= 3x(x + 3) + 2(x + 3)$

$\rm :\longmapsto\: {2x}^{2} + 2x + 3x + 3= {3x}^{2} + 9x + 2x + 6$

$\rm :\longmapsto\: {2x}^{2} + 5x + 3 = {3x}^{2} +11x + 6$

$\rm :\longmapsto\: {2x}^{2} + 5x + 3 - {3x}^{2} - 11x - 6 = 0$

$\rm :\longmapsto\: - {x}^{2} - 6x - 3 = 0$

$\rm :\longmapsto\: - ({x}^{2} + 6x + 3) = 0$

$\rm :\longmapsto\: {x}^{2} + 6x + 3 = 0$

Its a quadratic equation, whose solution is given by

$\rm :\longmapsto\:x = \dfrac{ - b \: \pm \: \sqrt{ {b}^{2} - 4ac} }{2a}$

Here,

a = 1

b = 6

c = 3

On substituting the values,

$\rm :\longmapsto\:x = \dfrac{ - 6 \: \pm \: \sqrt{ {6}^{2} - 4(1)(3)} }{2(1)}$

$\rm :\longmapsto\:x = \dfrac{ - 6 \: \pm \: \sqrt{ 36 - 12 }}{2}$

$\rm :\longmapsto\:x = \dfrac{ - 6 \: \pm \: \sqrt{ 24 }}{2}$

$\rm :\longmapsto\:x = \dfrac{ - 6 \: \pm \: \sqrt{2 \times 2 \times 6 }}{2}$

$\rm :\longmapsto\:x = \dfrac{ - 6 \: \pm \: 2\sqrt{ 6}}{2}$

$\rm :\longmapsto\:x = - 3 \: \pm \: \sqrt{6}$