Question

Solve for x:

$\frac{a}{x - a} + \frac{b}{x - b} = \frac{2c}{x - c}$

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Answer 1

$\huge{\mathfrak{Heya!!}}$

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Solution:-

$\frac{a}{x - a} + \frac{b}{x - b} = \frac{2c}{x - c} \\ \\ => ( \frac{a}{x - a} + 1) + ( \frac{b}{x - b} + 1) - ( \frac{2c}{x - c} + 2) = 0 \\ \\ = > \frac{a + x - a}{x - a} + \frac{b + x - b}{x - b} - \frac{2c + 2x - 2c}{x - c} = 0 \\ \\ = > \: x( \frac{1}{x - a} + \frac{1}{x - b} - \frac{2}{x - c} ) = 0 \\ \\ = > x.\frac{(x - b)( x - c) + (x - a)(x - c) - 2(x - a)(x - b)}{(x - a)(x - b)(x - c)} = 0 \\ \\ = > \: x( {x}^{2} - (b + c)x + bc + {x}^{2} + ac - 2 {x}^{2} + 2(a + b)x - 2ab) = 0 \\ \\ = > x((a + b - 2c)x + bc + ac - 2ab) = 0 \\ \\ = > x \: = 0 \: \: \: or \: (a + b - 2c)x + bc - 2ab = 0 \\ \\ x = 0 \: or \: x = \frac{2ab - bc - ac}{a + b - 2c}$
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Answer 2

Answer:

Value of x is 0 or ( 2ab - bc - ac ) / ( a + b - 2c )

Step-by-step explanation:

Given equation :

$\dfrac{a}{x-a}+\dfrac{b}{x-b} = \dfrac{2c}{x-c}$

Adding 2 on both sides,

$\implies \dfrac{a}{x-a}+\dfrac{b}{x-b} + 2= \dfrac{2c}{x-c} + 2 \\\\\\\implies \dfrac{a}{x-a}+1+\dfrac{b}{x-b}+1=\dfrac{2c+2(x-c)}{x-c}\\\\\\\implies \dfrac{a+x-a}{x-a}+\dfrac{b+x-b}{x-b}=\dfrac{2c+2x-2c}{x-c}\\\\\\\implies \dfrac{x}{x-a} +\dfrac{x}{x-b}=\dfrac{2x}{x-c}\\\\\\\implies x \bigg[ \dfrac{1}{x-a} + \dfrac{1}{x-b} \bigg] -x \bigg( \dfrac{2}{x-c}\bigg) = 0$

$\\\implies x \bigg[ \dfrac{1}{x-a} + \dfrac{1}{x-b}- \dfrac{2}{x-c}\bigg]= 0$

Now, x = 0  Or,

$\implies \dfrac{1}{x-a} +\dfrac{1}{x-b} - \dfrac{2}{x-c}=0$

$\implies \dfrac{(x-b)(x-c) + (x-a)(x-c)- 2(x-a)(x-b)}{(x-a)(x-b)(x-c)}=0\\\\\\\implies x^2 - ( b+c)x+bc+x^2 - (a+c)x + ac-2x^2+2(a+b)x-2ab=0\\\\\implies ( b+a-2c)x+bc+ac-2ab=0\\\\\implies x = \dfrac{2ab-bc-ac}{a+b-2c}$

Thus,

x = 0 or ( 2ab - bc - ac ) / ( a + b - 2c )