Math, asked by kabeertg, 11 days ago

solve for x :
 \frac{x  -  1}{x - 2}  +  \frac{x - 3}{x - 4}  = 3 \frac{1}{3}
;x is not equals to 2 and 4​

Answers

Answered by dhirajayush01
12

Answer:

x=5

Step-by-step explanation:

x-1/x-2 + x-3/x-4 = 10/3

x-1+x-3/x-2+x-4 = 10/3

x+x-1-3/x+x-2-4 = 10/3

2x-4/2x-6 = 10/3

2/3 = 10/3

10/3 × 3/2

5

x=5

Answered by mathdude500
18

Answer:

\qquad\qquad\boxed{ \bf{ \:x \:  = \:  5 \: , \:  \:  \dfrac{5}{2}  \: }} \\  \\

Step-by-step explanation:

Given equation is

\sf \: \dfrac{x - 1}{x - 2}  + \dfrac{x - 3}{x - 4}  = 3\dfrac{1}{3}  \\  \\

can be rewritten as

\sf \: \dfrac{x - 2 + 1}{x - 2}  + \dfrac{x - 4 + 1}{x - 4}  = \dfrac{10}{3}  \\  \\

\sf \: \dfrac{x - 2 }{x - 2}  + \dfrac{1}{x - 2}  + \dfrac{x - 4}{x - 4} + \dfrac{1}{x - 4}   = \dfrac{10}{3}  \\  \\

\sf \: 1  + \dfrac{1}{x - 2}  +1 + \dfrac{1}{x - 4}   = \dfrac{10}{3}  \\  \\

\sf \: 2  + \dfrac{1}{x - 2} + \dfrac{1}{x - 4}   = \dfrac{10}{3}  \\  \\

\sf \: \dfrac{1}{x - 2} + \dfrac{1}{x - 4}   = \dfrac{10}{3}  - 2   \\  \\

\sf \: \dfrac{x - 4 + x - 2}{(x - 2)(x - 4)}  = \dfrac{10 - 6}{3}   \\  \\

\sf \: \dfrac{2x - 6}{ {x}^{2}  - 4x - 2x + 8}  = \dfrac{4}{3}   \\  \\

\sf \: \dfrac{2(x - 3)}{ {x}^{2}  - 6x+ 8}  = \dfrac{4}{3}   \\  \\

\sf \: \dfrac{x - 3}{ {x}^{2}  - 6x+ 8}  = \dfrac{2}{3}   \\  \\

\sf \: 2( {x}^{2} - 6x + 8) = 3(x - 3) \\  \\

\sf \: 2{x}^{2} - 12x + 16 = 3x - 9 \\  \\

\sf \: 2{x}^{2} - 12x + 16  - 3x  +  9 = 0 \\  \\

\sf \: 2{x}^{2} - 15x + 25= 0 \\  \\

\sf \: 2{x}^{2} - 10x  - 5x+ 25= 0 \\  \\

\sf \: 2x(x - 5) - 5(x - 5) = 0 \\  \\

\sf \: (x - 5)(2x - 5) = 0 \\  \\

\sf \: x  - 5 = 0 \:  \:  \: or \:  \:  \: 2x - 5 = 0 \\  \\

\sf\implies  \: x = 5 \:  \:  \: or \:  \:  \: x = \dfrac{5}{2}  \\  \\

Hence,

\sf\implies  \:  \: \boxed{ \bf{ \:x \:  = \:  5 \: , \:  \:  \dfrac{5}{2}  \: }} \\  \\

\rule{190pt}{2pt}

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

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