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(D) Given equation: $\frac{1}{x-2}+\frac{1}{x+3}=\frac{7}{2x}$Multiplying both sides by the least common multiple $(LCM)$ of the denominators,which is

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$\left\{-\frac{14}{3}, 3\right\}$

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(D) Given equation: $\frac{1}{x-2}+\frac{1}{x+3}=\frac{7}{2x}$Multiplying both sides by the least common multiple $(LCM)$ of the denominators,which is $2x(x-2)(x+3)$:$2x(x+3) + 2x(x-2) = 7(x-2)(x+3)$Expanding the terms:$2x^2 + 6x + 2x^2 - 4x = 7(x^2 + x - 6)$$4x^2 + 2x = 7x^2 + 7x - 42$Rearranging the terms to form a standard quadratic equation $ax^2 + bx + c = 0$:$7x^2 - 4x^2 + 7x - 2x - 42 = 0$$3x^2 + 5x - 42 = 0$Factorizing the quadratic equation by splitting the middle term:$3x^2 + 14x - 9x - 42 = 0$$x(3x + 14) - 3(3x + 14) = 0$$(3x + 14)(x - 3) = 0$Setting each factor to zero:$3x + 14 = 0 \implies x = -\frac{14}{3}$$x - 3 = 0 \implies x = 3$Thus,the solution set is $\left\{-\frac{14}{3}, 3\right\}$.

Solve the following equation using the method of factorization and write its solution set: $\frac{1}{x-2}+\frac{1}{x+3}=\frac{7}{2x}$ $(x \neq 2, x \neq -3, x \neq 0)$

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