Solve the following equation using the method | vedclass

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Question

(A) Let $y = \frac{x+1}{x-3}$. Then the equation becomes $y + \frac{1}{y} = \frac{5}{2}$.Multiplying by $2y$,we get $2y^2 + 2 = 5y$,which simplifies t

Vedclass · Accepted Answer

$\{-5, 7\}$

Vedclass · Answer

(A) Let $y = \frac{x+1}{x-3}$. Then the equation becomes $y + \frac{1}{y} = \frac{5}{2}$.Multiplying by $2y$,we get $2y^2 + 2 = 5y$,which simplifies to $2y^2 - 5y + 2 = 0$.Factoring the quadratic equation: $2y^2 - 4y - y + 2 = 0 \implies 2y(y - 2) - 1(y - 2) = 0$.This gives $(2y - 1)(y - 2) = 0$,so $y = \frac{1}{2}$ or $y = 2$.Case $1$: $\frac{x+1}{x-3} = \frac{1}{2} \implies 2x + 2 = x - 3 \implies x = -5$.Case $2$: $\frac{x+1}{x-3} = 2 \implies x + 1 = 2x - 6 \implies x = 7$.Thus,the solution set is $\{-5, 7\}$.

Solve the following equation using the method of factorization and write its solution set: $\frac{x+1}{x-3}+\frac{x-3}{x+1}=\frac{5}{2}$

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