# Verify if the equation has any solutions 2x/(x+5)-x/(x-5)=50/(25-x^2) .

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have to determine if 2x/(x+5) - x/(x-5) = 50/(25-x^2) has any solutions.

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

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

=> 2x^2 - 10x - x^2 - 5x + 50 = 0

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

=> x^2 - 10x - 5x + 50 = 0

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

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

x = 5 and x = 10

But x = 5 makes x/(x-5) indeterminate

So the required solution of the equation is x = 10

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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

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

x1 = (15 + 5)/2

x1 = 10

x2 = (15-5)/2

x2 = 5

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

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