Solve \( 3-\frac{4}{x-3}=\frac{5}{x} \) \( x=\square \) or \( x=\square \)

To solve the equation \( 3 - \frac{4}{x-3} = \frac{5}{x} \), first, let's eliminate the fractions by finding a common denominator, which is \( x(x-3) \). Multiplying every term by \( x(x-3) \) gives: \( 3x(x-3) - 4x = 5(x-3) \) Expanding each term: \( 3x^2 - 9x - 4x = 5x - 15 \) Now simplify and combine like terms: \( 3x^2 - 13x + 15 = 0 \) Next, we can factor this quadratic equation. We need two numbers that multiply to \( 3 \times 15 = 45 \) and add to \( -13 \). Those numbers are \( -5 \) and \( -9 \). Thus, we can rewrite the equation as: \( 3x^2 - 5x - 9x + 15 = 0 \) Factoring by grouping gives us: \( (3x^2 - 5x) + (-9x + 15) = 0 \) Factoring further: \( x(3x - 5) - 3(3x - 5) = 0 \) This results in: \( (3x - 5)(x - 3) = 0 \) Setting each factor to zero gives: \( 3x - 5 = 0 \) or \( x - 3 = 0 \) Solving these yields: \( x = \frac{5}{3} \) or \( x = 3 \) However, since \( x = 3 \) would make the original expression undefined (due to division by zero), we discard that solution. The final answer is: \( x = \frac{5}{3} \) or \( x = \square \) (remaining solution is not applicable since \( x=3 \) is invalid). For clarity in completing the solution: \( x = \frac{5}{3} \) or \( x = \square \) (valid answer only is \( \frac{5}{3} \)).