Question

Solve for x:

$\frac{x - 4}{x - 5} + \frac{x - 6}{x - 7} = \frac{10}{3}$



Class 10


Quadratic Equation

Answer 1

Hey there !!


▶ Solve for x :-


$\bf \frac{x - 4}{x - 5} + \frac{x - 6}{x - 7} = \frac{10}{3} .$

▶ Solution :-

=> $\frac{x - 4}{x - 5} + \frac{x - 6}{x - 7} = \frac{10}{3} .$

$= > \frac{(x - 4)(x - 7) + (x - 6)(x - 5)}{(x - 5)(x - 7)} = \frac{10}{3} .$

$= > \frac{ {x}^{2} - 7x - 4x + 28 + {x}^{2} - 5x - 6x + 30}{ {x}^{2} - 7x - 5x + 35} = \frac{10}{3} .$


$= > \frac{2 {x}^{2} - 22x + 58 }{ {x}^{2} - 12x + 35 } = \frac{10}{3} .$


=> 3( 2x² - 22x + 58 ) = 10( x² - 12x + 35 ) .

=> 6x² - 66x + 174 = 10x² - 120x + 350.

=> 10x² - 6x² - 120x + 66x + 350 - 174 = 0.

=> 4x² - 54x + 176 = 0.

=> 2( 2x² - 27x + 88 ) = 0.

=> 2x² - 16x - 11x + 88 = 0.

=> 2x( x - 8 ) -11( x - 8 ) = 0.

=> 2x - 11 = 0 | x - 8 = 0.

=> x = 11/2. or x = 8.


✔✔ Hence, it is solved ✅✅.

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THANKS



#BeBrainly.

Answer 2

HEY THERE!!!


$\mathtt{ { \ \frac{x - 4}{x - 5} + \frac{x - 6}{x - 7} = \frac{10}{3}}} \\ \\ \\ \\ \sf \frac{(x - 4)(x - 7) + (x - 6)(x - 5)}{(x - 5)(x - 7)} = \frac{10}{3} \\ \\ \\ \sf \frac{ ({x}^{2} - 7x - 4x + 28) + ( {x}^{2} - 6x - 5x + 30) }{ {x}^{2} - 7x - 5x + 35} = \frac{10}{3} \tt\\ \\ \frac{( {x}^{2} - 11x + 28) + ({x}^{2} - 11x + 30 ) }{ {x }^{2} - 12x + 35} = \frac{10}{3} \\ \\ \\ \sf\frac{ {2x}^{2} - 22x + 58}{ {x}^{2} - 12x + 30} = \frac{10}{3} \\ \\ \\ \sf 3({2x}^{2} - 22x + 58) = 10( {x}^{2} - 12x + 35) \\ \\ \\ \sf {6x}^{2} - 66x + 174 = {10x}^{2} - 120x + 350 \\ \\ \\ {10x}^{2} - {6x}^{2} - 120x + 66x + 350 - 174 = 0 \\ \\ \\ \sf{4x}^{2} - 54x + 176 = 0 \\ \\ \\ \sf2( {2x}^{2} - 27x + 88) = 0 \\ \\ \\ \\ ( {2x}^{2} - 27x + 88) = 0 \\ \\ \\ {2x}^{2} - 16x - 11x + 88 \\ \\ \\ \sf2x(x - 8) - 11(x - 8) \\ \\ \\ \\ \tt (2x - 11)(x - 8) = 0 \\ \\ \\ (2x - 11) = 0 \\ \tt\\ 2x = 11 \\ \\ x = \frac{11}{2} \\ \\ \\ \\ \sf \:other \: \: root \: \: x = 8$
HEY THERE!!


Hence, Roots are 11/2 and 8 of this Quardratic Equation;