Question

"\frac{2x}{x+1} +\frac{3(x-1)}{x} =5" alt="\frac{2x}{x+1} +\frac{3(x-1)}{x} =5" align="absmiddle" class="latex-formula">

Answer 1

Answer:

$x=-3$

Step-by-step explanation:

Given equation:

$\dfrac{2x}{x+1}+\dfrac{3(x+1)}{x}=5$

Make the denominators the same:

$\implies \dfrac{2x}{x+1} \cdot\dfrac{x}{x}+\dfrac{3(x+1)}{x} \cdot \dfrac{x+1}{x+1}=5$

$\implies \dfrac{2x^2}{x(x+1)}+\dfrac{3(x+1)^2}{x(x+1)} =5$

Combine the fractions:

$\implies \dfrac{2x^2+3(x+1)^2}{x(x+1)} =5$

Multiply both sides by x(x+1):

$\implies 2x^2+3(x+1)^2 =5x(x+1)$

Expand the brackets:

$\implies 2x^2+3(x^2+2x+1) =5x^2+5x$

$\implies 2x^2+3x^2+6x+3 =5x^2+5x$

Combine like terms:

$\implies 5x^2+6x+3=5x^2+5x$

Subtract 5x² from both sides:

$\implies 6x+3=5x$

Subtract 5x from both sides:

$\implies x+3=0$

Subtract 3 from both sides:

$\implies x=-3$

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