Question
Can you please give me a simple way to understand and solve this equation, I have to show my work
X over X-5 - 2 over x+5 = 50 over x^2-25 thank you
X over X-5 - 2 over x+5 = 50 over x^2-25 thank you
Answers
GPT-4o mini
To solve the equation
\[
\frac{X}{X - 5} - \frac{2}{X + 5} = \frac{50}{X^2 - 25},
\]
we can start by recognizing that \(X^2 - 25\) can be factored. This is a difference of squares:
\[
X^2 - 25 = (X - 5)(X + 5).
\]
Thus, we can rewrite the equation as:
\[
\frac{X}{X - 5} - \frac{2}{X + 5} = \frac{50}{(X - 5)(X + 5)}.
\]
Now, to eliminate the fractions, we can multiply both sides of the equation by \((X - 5)(X + 5)\), the common denominator:
\[
(X - 5)(X + 5) \left(\frac{X}{X - 5} - \frac{2}{X + 5}\right) = 50.
\]
Distributing the left side gives:
\[
(X + 5)X - (X - 5)2 = 50.
\]
Now simplify both terms:
1. For the first term: \((X + 5)X = X^2 + 5X\).
2. For the second term: \((X - 5)2 = 2X - 10\).
Putting it all together:
\[
X^2 + 5X - (2X - 10) = 50,
\]
which simplifies to:
\[
X^2 + 5X - 2X + 10 = 50.
\]
Combining like terms results in:
\[
X^2 + 3X + 10 = 50.
\]
Now, to isolate terms, we subtract 50 from both sides:
\[
X^2 + 3X + 10 - 50 = 0,
\]
which simplifies to:
\[
X^2 + 3X - 40 = 0.
\]
Next, we can factor the quadratic equation \(X^2 + 3X - 40\):
We look for two numbers that multiply to \(-40\) and add to \(3\). These numbers are \(8\) and \(-5\). Thus, we can factor the equation as:
\[
(X + 8)(X - 5) = 0.
\]
Setting each factor to zero gives us:
1. \(X + 8 = 0\) → \(X = -8\)
2. \(X - 5 = 0\) → \(X = 5\)
However, we have to check if the solutions are valid. Since the original equation has restrictions due to the denominators, we note that \(X\) cannot equal \(5\) (because the denominator \(X - 5\) would be zero).
Thus, the only valid solution is:
\[
\boxed{-8}.
\]
\[
\frac{X}{X - 5} - \frac{2}{X + 5} = \frac{50}{X^2 - 25},
\]
we can start by recognizing that \(X^2 - 25\) can be factored. This is a difference of squares:
\[
X^2 - 25 = (X - 5)(X + 5).
\]
Thus, we can rewrite the equation as:
\[
\frac{X}{X - 5} - \frac{2}{X + 5} = \frac{50}{(X - 5)(X + 5)}.
\]
Now, to eliminate the fractions, we can multiply both sides of the equation by \((X - 5)(X + 5)\), the common denominator:
\[
(X - 5)(X + 5) \left(\frac{X}{X - 5} - \frac{2}{X + 5}\right) = 50.
\]
Distributing the left side gives:
\[
(X + 5)X - (X - 5)2 = 50.
\]
Now simplify both terms:
1. For the first term: \((X + 5)X = X^2 + 5X\).
2. For the second term: \((X - 5)2 = 2X - 10\).
Putting it all together:
\[
X^2 + 5X - (2X - 10) = 50,
\]
which simplifies to:
\[
X^2 + 5X - 2X + 10 = 50.
\]
Combining like terms results in:
\[
X^2 + 3X + 10 = 50.
\]
Now, to isolate terms, we subtract 50 from both sides:
\[
X^2 + 3X + 10 - 50 = 0,
\]
which simplifies to:
\[
X^2 + 3X - 40 = 0.
\]
Next, we can factor the quadratic equation \(X^2 + 3X - 40\):
We look for two numbers that multiply to \(-40\) and add to \(3\). These numbers are \(8\) and \(-5\). Thus, we can factor the equation as:
\[
(X + 8)(X - 5) = 0.
\]
Setting each factor to zero gives us:
1. \(X + 8 = 0\) → \(X = -8\)
2. \(X - 5 = 0\) → \(X = 5\)
However, we have to check if the solutions are valid. Since the original equation has restrictions due to the denominators, we note that \(X\) cannot equal \(5\) (because the denominator \(X - 5\) would be zero).
Thus, the only valid solution is:
\[
\boxed{-8}.
\]