Question

$\frac{2x}{x - 4} + \frac{2x - 5}{x - 3} = \frac{25}{3}$
solve for x. ​

Answer 1

$\sf\huge\bold{\underline{\underline{{Solution}}}}$

$\sf = > \dfrac{2x}{x - 4} + \dfrac{2x - 5}{x - 3} = \dfrac{25}{3}$

Taking LCM of x-4 & x-3

$\sf = > \dfrac{(x - 3)2x + (2x - 5)(x - 4)}{(x - 4)(x - 3)} = \dfrac{25}{3}$

$\sf = > \dfrac{2 {x}^{2} - 6x + 2x(x - 4) - 5(x - 4)}{x(x - 3) - 4(x - 3)} = \dfrac{25}{3}$

$\sf = > \dfrac{2 {x}^{2} - 6x + 2 {x}^{2} - 8x - 5x + 20}{ {x}^{2} - 3x - 4x + 12} = \dfrac{25}{3}$

Making like terms in the numerator and in the denominator:

$\sf = > \dfrac{2 {x}^{2} + 2 {x}^{2} - 6x - 8x - 5x + 20 }{ {x}^{2} - 3x - 4x + 12 } = \dfrac{25}{3}$

$\sf = > \dfrac{4 {x}^{2} - 19x + 20 }{ {x}^{2} - 7x + 12} = \dfrac{25}{3}$

Cross multiply to both sides:

➙3(4x²-19x+20)=25(x²-7x+12)

➙12x²-57x+60=25x²-175x+300

➙25x²-12x²-175x+57x+300-60

➙13x²-118x+240

Now,it is in the form of a quadratic equation so,we factorise it .

We use quadratic formula for factorising:

a= 13 ,b= -118 & c= 240

x=-b±√b²-4ac÷2a

➙x=-(-118)±√(-118)²-4(13)(240)/2(13)

➙x=118±√13924-12480/26

➙x=118±√1444/26

➙x=118±38/26

➙x=156/26 or x = 80/26

➙x=6 or x = 40/13

➙x=6 or x= 3.07

Therefore,value of x is 6 or 3.07