c) \( x+1=\frac{15}{x-1} \)

To solve the equation \( x + 1 = \frac{15}{x - 1} \), start by eliminating the fraction. Multiply both sides by \( x - 1 \) (noting that \( x \neq 1 \)), giving you: \[ (x + 1)(x - 1) = 15 \] This expands to: \[ x^2 - 1 = 15 \] Then, move 15 to the left side: \[ x^2 - 16 = 0 \] Factoring gives \( (x - 4)(x + 4) = 0 \), so the solutions are \( x = 4 \) and \( x = -4 \). However, remember to check the solutions in the original equation to ensure they don’t lead to division by zero. Here, neither \( x = 4 \) nor \( x = -4 \) makes the denominator zero. Hence, both are valid solutions! Now let’s talk about how these kinds of equations can appear in real life—like in physics where we solve for unknowns in motion equations. Oftentimes, these variables can represent objects' velocities or displacements under certain conditions. It’s just like putting a puzzle together where each piece represents a part of the scenario you’re analyzing! And speaking of problem-solving, a common mistake people make is forgetting to check for undefined values (like our \( x - 1 \) denominator). Always remember to ask yourself: does my solution make the original equation unusable? If it does, it’s time for a re-evaluation!