Question

]if f(x) = x+3 / x+2 , prove that x= (2f(x) - 3) / (1-f(x))

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{f(x)=\dfrac{x+3}{x+2}}$

$\underline{\textbf{To prove:}}$

$\mathsf{x=\dfrac{2\,f(x)-3}{1-f(x)}}$

$\underline{\textbf{Solution:}}$

$\mathsf{Consider,}$

$\mathsf{\dfrac{2\,f(x)-3}{1-f(x)}}$

$\mathsf{=\dfrac{2\left(\dfrac{x+3}{x+2}\right)-3}{1-\left(\dfrac{x+3}{x+2}\right)}}$

$\mathsf{=\dfrac{\dfrac{2x+6-3(x+2)}{x+2}}{\dfrac{x+2-x-3}{x+2}}}$

$\mathsf{=\dfrac{\dfrac{2x+6-3x-6}{x+2}}{\dfrac{2-3}{x+2}}}$

$\mathsf{=\dfrac{\dfrac{-x}{x+2}}{\dfrac{-1}{x+2}}}$

$\mathsf{=\dfrac{-x}{-1}}$

$\mathsf{=x}$

$\implies\boxed{\mathsf{\dfrac{2\,f(x)-3}{1-f(x)}=x}}$

Answer 2

SOLUTION

GIVEN

$\displaystyle \sf{f(x) = \frac{x + 3}{x + 2} }$

TO PROVE

$\displaystyle \sf{ x = \frac{ 2f(x) - 3}{1 - f(x) } }$

PROOF

$\displaystyle \sf{f(x) = \frac{x + 3}{x + 2} }$

Cross multiplication gives

$\displaystyle \sf{(x + 2)f(x) = x + 3 }$

$\displaystyle \sf{ \implies \: x f(x)+ 2f(x) = x + 3 }$

$\displaystyle \sf{ \implies \: x f(x) - x = 3 - 2f(x) }$

$\displaystyle \sf{ \implies \: x \bigg[f(x) - 1 \bigg]= 3 - 2f(x) }$

$\displaystyle \sf{ \implies \: x = \frac{3 - 2f(x)}{f(x) - 1 } }$

Multiplying numerator and denominator both by - 1 we get

$\displaystyle \sf{ \implies \: x = \frac{ 2f(x) - 3}{1 - f(x) } }$

Hence proved

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