Solve the equation: 2(3x-5/x+2)-5(x+2/3x-5)=3 , where x is not equal t

1.	Solve the equation: 2(3x-5/x+2)-5(x+2/3x-5)=3 , where x is not equal to -2 and 5/3
Answer» $\large\underline{\sf{Solution-}}$ The GIVEN equation is $\bf :\longmapsto\:2\bigg(\dfrac{3x - 5}{x + 2} \bigg) - 5\bigg(\dfrac{x + 2}{3x - 5} \bigg) = 3 - - - (1)$ To SOLVE this equation, Let we assume that, $\red{\rm :\longmapsto\:\dfrac{3x - 5}{x + 2} = y - - - (2)}$ So equation (1) can be rewritten as $\rm :\longmapsto\:2y - \dfrac{5}{y} = 3$ $\rm :\longmapsto\: \dfrac{ {2y}^{2} - 5}{y} = 3$ $\rm :\longmapsto\: {2y}^{2} - 5 = 3y$ $\rm :\longmapsto\: {2y}^{2} - 3y - 5 = 0$ $\rm :\longmapsto\: {2y}^{2} - 5y + 2y - 5 = 0$ $\rm :\longmapsto\:y(2y - 5) + 1(2y - 5) = 0$ $\rm :\longmapsto\:(2y - 5) (y+ 1) = 0$ $\rm :\longmapsto\:2y - 5 = 0 \: \: \: or \: \: \: y+ 1= 0$ $\bf\implies \:y = \dfrac{5}{2} \: \: \: or \: \: \: y = - \: 1$ So, Case :- 1 $\red{\rm :\longmapsto\:y = \dfrac{5}{2} }$ $\rm :\longmapsto\:\dfrac{3x - 5}{x + 2} = \dfrac{5}{2}$ $\rm :\longmapsto\:6x - 10 = 5x + 10$ $\rm :\longmapsto\:6x - 5x = 10 + 10$ $\bf :\longmapsto\:x = 20$ Now, Case :- 2 $\red{\rm :\longmapsto\:y \: = \: - \: 1}$ $\rm :\longmapsto\:\dfrac{3x - 5}{x + 2} = - \: 1$ $\rm :\longmapsto\:3x - 5 = - x - 2$ $\rm :\longmapsto\:3x + x = 5- 2$ $\rm :\longmapsto\:4x = 3$ $\bf\implies \:x = \dfrac{3}{4}$ Additional Information :- Nature of ROOTS :- Let US consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation. If Discriminant, D > 0, then roots of the equation are real and unequal. If Discriminant, D = 0, then roots of the equation are real and equal. If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary. Where, Discriminant, D = b² - 4ac

Solve the equation: 2(3x-5/x+2)-5(x+2/3x-5)=3 , where x is not equal to -2 and 5/3

Answer»

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

The GIVEN equation is

$\bf :\longmapsto\:2\bigg(\dfrac{3x - 5}{x + 2} \bigg) - 5\bigg(\dfrac{x + 2}{3x - 5} \bigg) = 3 - - - (1)$

To SOLVE this equation,

Let we assume that,

$\red{\rm :\longmapsto\:\dfrac{3x - 5}{x + 2} = y - - - (2)}$

So equation (1) can be rewritten as

$\rm :\longmapsto\:2y - \dfrac{5}{y} = 3$

$\rm :\longmapsto\: \dfrac{ {2y}^{2} - 5}{y} = 3$

$\rm :\longmapsto\: {2y}^{2} - 5 = 3y$

$\rm :\longmapsto\: {2y}^{2} - 3y - 5 = 0$

$\rm :\longmapsto\: {2y}^{2} - 5y + 2y - 5 = 0$

$\rm :\longmapsto\:y(2y - 5) + 1(2y - 5) = 0$

$\rm :\longmapsto\:(2y - 5) (y+ 1) = 0$

$\rm :\longmapsto\:2y - 5 = 0 \: \: \: or \: \: \: y+ 1= 0$

$\bf\implies \:y = \dfrac{5}{2} \: \: \: or \: \: \: y = - \: 1$

So,

Case :- 1

$\red{\rm :\longmapsto\:y = \dfrac{5}{2} }$

$\rm :\longmapsto\:\dfrac{3x - 5}{x + 2} = \dfrac{5}{2}$

$\rm :\longmapsto\:6x - 10 = 5x + 10$

$\rm :\longmapsto\:6x - 5x = 10 + 10$

$\bf :\longmapsto\:x = 20$

Now,

Case :- 2

$\red{\rm :\longmapsto\:y \: = \: - \: 1}$

$\rm :\longmapsto\:\dfrac{3x - 5}{x + 2} = - \: 1$

$\rm :\longmapsto\:3x - 5 = - x - 2$

$\rm :\longmapsto\:3x + x = 5- 2$

$\rm :\longmapsto\:4x = 3$

$\bf\implies \:x = \dfrac{3}{4}$

Additional Information :-

Nature of ROOTS :-

Let US consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

If Discriminant, D > 0, then roots of the equation are real and unequal.

If Discriminant, D = 0, then roots of the equation are real and equal.

If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.