6/x^2-25 - 1/x^2-5x =2/x^2-5x solve for the unknown write down restrictions
To solve for the unknown in the equation:
6/(x^2-25) - 1/(x^2-5x) = 2/(x^2-5x)
First, we need to factor the denominators:
6/[(x-5)(x+5)] - 1/[x(x-5)] = 2/[x(x-5)]
Next, we need to find a common denominator:
[x(x-5)*(x+5)]
Multiplying each fraction by the appropriate factor to get the common denominator, we get:
6x/(x(x-5)(x+5)) - (x+5)/(x(x-5)(x+5)) = 2(x-5)/(x(x-5)(x+5))
Simplifying the left-hand side, we get:
(6x - (x+5))/(x(x-5)(x+5)) = (2x-10)/(x(x-5)(x+5))
Combining like terms on the left-hand side, we get:
(5x-5)/(x(x-5)(x+5)) = (2x-10)/(x(x-5)(x+5))
Multiplying both sides by the common denominator, we get:
5x-5 = 2x-10
Solving for x, we get:
x = -1
However, we need to check if this solution is valid. Since we have denominators in the original equation and in our factoring, we need to check for any restrictions on x that would make those denominators equal to zero and make the equation undefined. In this case, we have:
x^2 - 25 = 0 (the denominator of the first fraction)
x^2 - 5x = 0 (the denominator of the second and third fractions)
Factoring these equations, we get:
(x+5)(x-5) = 0 (the denominator of the first fraction)
x(x-5) = 0 (the denominator of the second and third fractions)
Solving for x in each equation, we get:
x = 5, -5 (for the first denominator)
x = 0, 5 (for the second and third denominators)
We see that x = 5 and x = -5 are not valid solutions because they make the denominators in the original equation equal to zero and the equation undefined. We also see that x = 0 is not a valid solution because it makes the second and third denominators equal to zero and the equation undefined.
Therefore, the only valid solution is:
x = -1.