Solve the following equation using the method | vedclass

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Question

(C) Given equation: $\frac{x}{x+1} + \frac{x+1}{x} = \frac{41}{20}$.Let $y = \frac{x}{x+1}$,then the equation becomes $y + \frac{1}{y} = \frac{41}{20}

Vedclass · Accepted Answer

$\{4, -5/4\}$

Vedclass · Answer

(C) Given equation: $\frac{x}{x+1} + \frac{x+1}{x} = \frac{41}{20}$.Let $y = \frac{x}{x+1}$,then the equation becomes $y + \frac{1}{y} = \frac{41}{20}$.Multiplying by $20y$,we get $20y^2 + 20 = 41y$,which rearranges to $20y^2 - 41y + 20 = 0$.Factorizing the quadratic: $20y^2 - 16y - 25y + 20 = 0 \Rightarrow 4y(5y - 4) - 5(5y - 4) = 0$.So,$(4y - 5)(5y - 4) = 0$,giving $y = 5/4$ or $y = 4/5$.Case $1$: $\frac{x}{x+1} = \frac{5}{4} \Rightarrow 4x = 5x + 5 \Rightarrow x = -5$.Case $2$: $\frac{x}{x+1} = \frac{4}{5} \Rightarrow 5x = 4x + 4 \Rightarrow x = 4$.Thus,the solution set is $\{4, -5\}$.

Solve the following equation using the method of factorization and write its solution set: $\frac{x}{x+1} + \frac{x+1}{x} = \frac{41}{20}$

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