Solve for $$ x $$ $$ \frac{x-4}{x}+\frac{3}{3}=\frac{-3}{x} $$ $$ x=\square $$ $$ \rightarrow $$ Next Question

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$$ x = \frac{1}{2} $$. To solve the equation $$ \frac{x-4}{x} + \frac{3}{3} = \frac{-3}{x}, $$ we first simplify the equation. The term $$\frac{3}{3}$$ simplifies to $$1$$. Thus, we can rewrite the equation as: $$ \frac{x-4}{x} + 1 = \frac{-3}{x}. $$ Next, we can eliminate the fractions by multiplying the entire equation by $$x$$ (assuming $$x \neq 0$$): $$ x \cdot \left(\frac{x-4}{x}\right) + x \cdot 1 = x \cdot \left(\frac{-3}{x}\right). $$ This simplifies to: $$ x - 4 + x = -3. $$ Combining like terms gives: $$ 2x - 4 = -3. $$ Next, we add $$4$$ to both sides: $$ 2x = -3 + 4, $$ which simplifies to: $$ 2x = 1. $$ Now, we divide both sides by $$2$$: $$ x = \frac{1}{2}. $$ Thus, the solution is $$ x = \frac{1}{2}. $$ So, the final answer is: $$ x = \frac{1}{2}. $$

Solve for \( x \) \( \frac{x-4}{x}+\frac{3}{3}=\frac{-3}{x} \) \( x=\square \) \( \rightarrow \) Next Question

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Solve for \( x \) \( \frac{x-4}{x}+\frac{3}{3}=\frac{-3}{x} \) \( x=\square \) \( \rightarrow \) Next Question

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