Question
LaTeX: \frac{x}{x+5}=\frac{50}{x^2-25}+6
Answers
GPT-4o mini
To solve the equation
\[
\frac{x}{x+5} = \frac{50}{x^2 - 25} + 6,
\]
we start by noticing that \(x^2 - 25\) is a difference of squares and can be factored as
\[
x^2 - 25 = (x - 5)(x + 5).
\]
This means we can rewrite the equation as:
\[
\frac{x}{x + 5} = \frac{50}{(x - 5)(x + 5)} + 6.
\]
To eliminate the fractions, we can multiply both sides of the equation by \((x - 5)(x + 5)\):
\[
x(x - 5) = 50 + 6(x - 5)(x + 5).
\]
Now we distribute on the right side:
\[
x(x - 5) = 50 + 6(x^2 - 25).
\]
This simplifies to:
\[
x^2 - 5x = 50 + 6x^2 - 150.
\]
Combining like terms on the right side gives:
\[
x^2 - 5x = 6x^2 - 100.
\]
Next, we move all terms to one side of the equation:
\[
x^2 - 5x - 6x^2 + 100 = 0,
\]
which simplifies to:
\[
-5x^2 - 5x + 100 = 0.
\]
We can divide the whole equation by -5:
\[
x^2 + x - 20 = 0.
\]
Now, we can factor this quadratic:
\[
(x + 5)(x - 4) = 0.
\]
Setting each factor equal to zero gives us:
\[
x + 5 = 0 \quad \Rightarrow \quad x = -5,
\]
\[
x - 4 = 0 \quad \Rightarrow \quad x = 4.
\]
Now we check if these values lead to any restrictions due to the original equation's denominators.
Substituting back into the denominators:
- For \(x = -5\), the expression \(\frac{x}{x + 5}\) becomes undefined, so \(x = -5\) is not valid.
- For \(x = 4\), it's valid since \(x + 5 \neq 0\) and \(x - 5 \neq 0\).
Thus, the only valid solution is
\[
\boxed{4}.
\]
\[
\frac{x}{x+5} = \frac{50}{x^2 - 25} + 6,
\]
we start by noticing that \(x^2 - 25\) is a difference of squares and can be factored as
\[
x^2 - 25 = (x - 5)(x + 5).
\]
This means we can rewrite the equation as:
\[
\frac{x}{x + 5} = \frac{50}{(x - 5)(x + 5)} + 6.
\]
To eliminate the fractions, we can multiply both sides of the equation by \((x - 5)(x + 5)\):
\[
x(x - 5) = 50 + 6(x - 5)(x + 5).
\]
Now we distribute on the right side:
\[
x(x - 5) = 50 + 6(x^2 - 25).
\]
This simplifies to:
\[
x^2 - 5x = 50 + 6x^2 - 150.
\]
Combining like terms on the right side gives:
\[
x^2 - 5x = 6x^2 - 100.
\]
Next, we move all terms to one side of the equation:
\[
x^2 - 5x - 6x^2 + 100 = 0,
\]
which simplifies to:
\[
-5x^2 - 5x + 100 = 0.
\]
We can divide the whole equation by -5:
\[
x^2 + x - 20 = 0.
\]
Now, we can factor this quadratic:
\[
(x + 5)(x - 4) = 0.
\]
Setting each factor equal to zero gives us:
\[
x + 5 = 0 \quad \Rightarrow \quad x = -5,
\]
\[
x - 4 = 0 \quad \Rightarrow \quad x = 4.
\]
Now we check if these values lead to any restrictions due to the original equation's denominators.
Substituting back into the denominators:
- For \(x = -5\), the expression \(\frac{x}{x + 5}\) becomes undefined, so \(x = -5\) is not valid.
- For \(x = 4\), it's valid since \(x + 5 \neq 0\) and \(x - 5 \neq 0\).
Thus, the only valid solution is
\[
\boxed{4}.
\]