Question

LaTeX: \frac{x}{x+7}-8=\frac{98}{x^2-49}

Answer 1

To solve the equation

\[
\frac{x}{x+7}-8=\frac{98}{x^2-49},
\]

we can first simplify the right-hand side. Note that \(x^2 - 49\) can be factored as

\[
x^2 - 49 = (x - 7)(x + 7).
\]

Substituting this into our equation gives:

\[
\frac{x}{x+7}-8=\frac{98}{(x - 7)(x + 7)}.
\]

Next, we can eliminate the fractions by finding a common denominator. The common denominator is \((x + 7)(x - 7)\):

Multiply both sides by \((x + 7)(x - 7)\):

\[
\left(\frac{x}{x+7}-8\right)(x + 7)(x - 7) = 98.
\]

Distributing on the left side:

\[
x(x - 7) - 8(x - 7)(x + 7) = 98.
\]

Now simplify each term:

1. \( x(x - 7) = x^2 - 7x \)
2. \( (x - 7)(x + 7) = x^2 - 49 \), therefore:
\( -8(x^2 - 49) = -8x^2 + 392 \)

Putting it all together results in:

\[
x^2 - 7x - 8x^2 + 392 = 98.
\]

Combine like terms:

\[
-7x^2 - 7x + 392 = 98.
\]

Next, subtract \(98\) from both sides:

\[
-7x^2 - 7x + 294 = 0.
\]

To make the coefficients easier to work with, divide everything by \(-7\):

\[
x^2 + x - 42 = 0.
\]

Now, we can factor this quadratic equation:

\[
(x + 7)(x - 6) = 0.
\]

Setting each factor equal to zero gives us:

\[
x + 7 = 0 \quad \Rightarrow \quad x = -7,
\]
\[
x - 6 = 0 \quad \Rightarrow \quad x = 6.
\]

We need to check for any restrictions. The original equation has a denominator of \(x^2 - 49\), which will be undefined for \(x = 7\) and \(x = -7\). Thus, we discard \(x = -7\) as a solution.

The solution to the equation is:

\[
\boxed{6}.
\]