# more than a questioni have to find the number of solutions and what are solutions of equation 2x/(x+5)=50/(25-x^2)+x/(x-5)

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The equation to be solved is 2x/(x+5)=50/(25-x^2)+x/(x-5)

2x/(x+5)=50/(25-x^2)+x/(x-5)

=> 2x/(x+5) - x/(x-5)=50/(25-x^2)

=> 2x(x - 5) + x(x + 5) = 50

=> 2x^2 - 10x + x^2 + 5x = 50

=> 3x^2 - 5x - 50 = 0

=> 3x^2 - 15x + 10x - 50 = 0

=> 3x(x - 5) + 10(x - 5) = 0

=> (3x + 10)(x - 5)=0

=> x = -10/3 and x = 5

There are two solutions of the equation x = -10/3 and x = 5

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First, we'll impose the constraints of existence of the fractions. All denominators have to be different from zero, for the fractions to exist.

x + 5 different from 0 => x different from -5

x - 5 different from 0 => x different from 5

The solutions of the equation can have any real value, except the values {-5 ; 5}.

We'll solve the equation:

2x(x-5) - x(x+5) = 50

2x^2 - 10x - x^2 - 5x = 50

We'll combine like terms:

x^2 - 15x - 50 = 0

We'll apply quadratic formula:

x1 = [15+sqrt(225 - 200)]/2

x1 = (15 + 5)/2

x1 = 10

x2 = (15-5)/2

x2 = 5

Since x2 = 5 is an excepted value, we'll reject this solution.

### The equation will have only one real solution: x = 10.

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