Solve the following equation using the method | vedclass

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(D) Let $y = \frac{x}{1-x}$. Then the equation becomes $y + \frac{1}{y} = \frac{13}{6}$.Multiplying by $6y$,we get $6y^2 + 6 = 13y$,which simplifies t

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$\{\frac{2}{3}, \frac{3}{2}\}$

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(D) Let $y = \frac{x}{1-x}$. Then the equation becomes $y + \frac{1}{y} = \frac{13}{6}$.Multiplying by $6y$,we get $6y^2 + 6 = 13y$,which simplifies to $6y^2 - 13y + 6 = 0$.Factoring the quadratic equation: $6y^2 - 9y - 4y + 6 = 0 \implies 3y(2y - 3) - 2(2y - 3) = 0$.This gives $(3y - 2)(2y - 3) = 0$,so $y = \frac{2}{3}$ or $y = \frac{3}{2}$.Case $1$: $\frac{x}{1-x} = \frac{2}{3} \implies 3x = 2 - 2x \implies 5x = 2 \implies x = \frac{2}{5}$.Case $2$: $\frac{x}{1-x} = \frac{3}{2} \implies 2x = 3 - 3x \implies 5x = 3 \implies x = \frac{3}{5}$.Thus,the solution set is $\{\frac{2}{5}, \frac{3}{5}\}$.

Solve the following equation using the method of factorization and write its solution set: $\frac{x}{1-x} + \frac{1-x}{x} = \frac{13}{6}$

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