Question

reduce the following equation into standard quadratic equation formed​

Answer 1

We have

$\sf \to \: \dfrac{x - 2}{x - 3} + \dfrac{x - 4}{x - 5} = \dfrac{10}{3}$

To Find

Reduce this equation into standard quadratic equation

Now Take

$\sf \to \: \dfrac{x - 2}{x - 3} + \dfrac{x - 4}{x - 5} = \dfrac{10}{3}$

Take LCM on LHS

$\sf \to \: \dfrac{(x - 2)(x - 5) + (x - 4)(x - 3)}{(x - 3)(x - 5)} = \dfrac{10}{3}$

$\sf \to \: \dfrac{ {x }^{2} - 5x - 2x + 10 + {x}^{2} - 3x - 4x + 12 }{ {x}^{2} - 5x - 3x + 15} = \dfrac{10}{3}$

$\sf \to \: \dfrac{ {x }^{2} - 7x + 10 + {x}^{2} - 7x + 12 }{ {x}^{2} -8x + 15} = \dfrac{10}{3}$

$\sf \to \: \dfrac{2 {x }^{2} - 14x + 22 }{ {x}^{2} -8x + 15} = \dfrac{10}{3}$

$\sf \to \: \dfrac{2( {x }^{2} - 7x + 11)}{ {x}^{2} -8x + 15} = \dfrac{10}{3}$

$\sf \to \: \dfrac{{x }^{2} - 7x + 11}{ {x}^{2} -8x + 15} = \dfrac{5}{3}$

Using Cross multiplication method

$\sf \to \: 3( {x}^{2} - 7x + 11) = 5( {x}^{2} - 8x + 15)$

$\sf \to \: 3 {x}^{2} - 21x + 33 = 5 {x}^{2} - 40x + 75$

$\sf \to \: 5 {x}^{2} - 3 {x}^{2} - 40x + 21x + 75 - 33 = 0$

$\sf \to \: 2 {x}^{2} - 19x + 42 = 0$

Now Find Value of x

$\sf \to \: 2 {x}^{2} - 19x + 42 = 0$

Factories the equation, we get

$\sf \to \: 2 {x}^{2} - 12x - 7x + 42 = 0$

$\sf \to \: 2x( {x}^{} - 6) - 7(x - 6) = 0$

$\sf \to \: (2x - 7)(x - 6) = 0$

$\sf \to \: x = \dfrac{7}{2} \: and \: x = 6$

Answer 2

Given Equation:-

$\\ \sf{ \bold{ \frac{x - 2}{x - 3} + \frac{x - 4}{x - 5} = \frac{10}{3} }}$

To Find:-

$\\ \sf{ \rightarrow{The \: value \: of \: \bold{x}}}$

♡ Solution:-

$\\ \sf{ \frac{x - 2}{x - 3} + \frac{x - 4}{x - 5} = \frac{10}{3} }$

$\\ \sf{ \implies{ \frac{(x - 2)(x - 5) + (x - 4)(x - 3)}{(x - 3)(x - 5)} = \frac{10}{3} }}$

$\\ \sf{ \implies{ \frac{x {}^{2} - 5x - 2x + 10 + {x}^{2} - 3x - 4x + 12 }{ {x}^{2} - 5x - 3x + 15 } = \frac{10}{3} }}$

$\\ \sf{ \implies{ \frac{2 {x}^{2} - 14x + 22 }{ {x}^{2} - 8x + 15 } = \frac{10}{3} }}$

$\\ \sf{ \implies{ \frac{2( {x}^{2} - 7x + 11 )}{ {x}^{2} - 8x + 15} = \frac{10}{3} }}$

$\\ \sf{ \implies{ \frac{ {x}^{2} - 7x + 11 }{ {x}^{2} - 8x + 15} = \frac{5}{3} }}$

$\\ \sf{ \implies{3 {x}^{2} - 21x + 33 = 5 {x}^{2} - 40x + 75}}$

$\\ \sf{ \implies{ 5 {x}^{2} - 40x + 75 - 3 {x}^{2} + 21x - 33 = 0}}$

$\\ \sf{ \implies{ 2{x}^{2} - 19x + 42= 0}}$

$\\ \sf{ \implies{2x {}^{2} - 12x - 7x + 42 = 0}}$

$\\ \sf{ \implies{2x(x - 6) - 7(x - 6) = 0}}$

$\\ \sf{ \implies{(x - 6)(2x - 7) = 0}}$

Hence,

$\sf{ \: \: x - 6 = 0} \\ \therefore \sf{ \orange{x = 6}}$

Again,

$\sf{ \: \: 2x - 7 = 0} \\ \sf{ \therefore{ \orange{x = \frac{7}{2} }}}$

❖Hence, the answer is:-

$\sf{ \bold{x}} = 6,\: \frac{7}{2}$