Simplifying each side of the equation results in \( x^{2}-3 x-4=x^{2}-5 x+6 \) Find the solution: \( \frac{x+2}{3 x}-\frac{1}{x-2}=\frac{x-3}{3 x} \)
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The Deep Dive
First, let’s simplify the original equation: \( x^{2}-3x-4=x^{2}-5x+6 \). Subtract \( x^2 \) from both sides to get: \(-3x - 4 = -5x + 6\). Now, add \( 5x \) to both sides: \(-3x + 5x = 6 + 4\), which simplifies to: \(2x = 10\). Dividing both sides by 2 gives: \(x = 5\). Now let’s solve the equation: \( \frac{x+2}{3x}-\frac{1}{x-2}=\frac{x-3}{3x} \). First, get a common denominator, which is \(3x(x-2)\): \( \frac{(x+2)(x-2)}{3x(x-2)} - \frac{3x}{3x(x-2)} = \frac{(x-3)(x-2)}{3x(x-2)} \). Now multiplying both sides by \(3x(x-2)\) (assuming \(x \neq 0\) and \(x \neq 2\)): \((x+2)(x-2) - 3x = (x-3)(x-2)\). Expanding both sides: \(x^2 - 4 - 3x = x^2 - 5x + 6\). Then, we can simplify: \(-4 + 3x = -5x + 6\) which becomes: \(8x = 10\). Thus: \(x = \frac{10}{8} = \frac{5}{4}\). The solutions are \(x = 5\) and \(x = \frac{5}{4}\).
