Simplify each expression. \( \frac{14 x^{2}-16 x}{x+6} \cdot \frac{x+6}{70 x-80} \)

To simplify the expression \( \frac{14 x^{2}-16 x}{x+6} \cdot \frac{x+6}{70 x-80} \), we can start by canceling out the common factors. First, rewrite the expression, noting that \(70 x - 80\) can be factored out: \[ \frac{14x^2 - 16x}{x+6} \cdot \frac{x+6}{70x - 80} = \frac{14x(x - \frac{16}{14})}{x+6} \cdot \frac{x+6}{10(7x - 8)} \] This helps us see more clearly what can be canceled. The \(x + 6\) terms in the numerator and denominator cancel out: \[ \frac{14x(x - \frac{8}{7})}{10(7x - 8)} \] Next, we can further simplify \( \frac{14x}{10} \), which reduces to \( \frac{7x}{5} \): So now we have: \[ \frac{7x(x - \frac{8}{7})}{5(7x - 8)} \] Now we can also note that \(x - \frac{8}{7}\) is the same as \(\frac{7x - 8}{7}\): Putting it all together, we can rewrite it as: \[ \frac{7x \cdot \frac{7x - 8}{7}}{5(7x - 8)} \] The \(7x - 8\) terms can cancel too: Finally, we get: \[ \frac{x}{5} \] So the simplified expression is \( \frac{x}{5} \).