What is the answer to a/((x-a))+b/((x-b))=((2c))/((x-c)) ?

The answer to a/((x-a))+b/((x-b))=((2c))/((x-c)) is x=0,x=-(-2ab+ac+bc)/(a+b-2c);\quad a+b-2c\ne 0

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a/((x-a))+b/((x-b))=((2c))/((x-c))

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

x = 0, x = - \frac{- 2 ab + a c + b c}{a + b - 2 c}; a + b - 2 c \neq = 0

Solution steps

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

Multiply by LCM

a (x - b) (x - c) + b (x - a) (x - c) = 2 c (x - a) (x - b)

Solve

a (x - b) (x - c) + b (x - a) (x - c) = 2 c (x - a) (x - b) : x = 0, x = - \frac{- 2 ab + a c + b c}{a + b - 2 c}; a + b - 2 c \neq = 0

x = 0, x = - \frac{- 2 ab + a c + b c}{a + b - 2 c}; a + b - 2 c \neq = 0

Verify Solutions

Find undefined (singularity) points:

x = c

Combine undefined points with solutions:

x = 0, x = - \frac{- 2 ab + a c + b c}{a + b - 2 c}; a + b - 2 c \neq = 0

2 (\frac{2 x - 1}{x + 3}) - 3 (\frac{x + 3}{2 x - 1}) = 5

x, \frac{1}{x} + \frac{1}{y} = 1

2 (\frac{x ^{2} + 1}{( x ^{2} )}) - 3 (\frac{x - 1}{x}) - 4 = 0

\frac{( 2 ^{n} + 2 ^{n - 1} )}{( 2 ^{n + 1} - 2 ^{n} )} = \frac{3}{2}

\frac{31}{2 x} = 56

What is the answer to a/((x-a))+b/((x-b))=((2c))/((x-c)) ?
The answer to a/((x-a))+b/((x-b))=((2c))/((x-c)) is x=0,x=-(-2ab+ac+bc)/(a+b-2c);\quad a+b-2c\ne 0