Question

If $\dfrac{\left( 2x+3 \right)\left( 3x-4 \right)}{\left( x-1 \right)\left( 4x+5 \right)}=f\left( x \right)-\dfrac{5}{9\left( x-1 \right)}+\dfrac{31}{18\left( 4x+5 \right)}$ ,then $f\left( x \right)=$

Answer 1

Hint: From the above question we have to find the $f\left( x \right)$ from $\dfrac{\left( 2x+3 \right)\left( 3x-4 \right)}{\left( x-1 \right)\left( 4x+5 \right)}=f\left( x \right)-\dfrac{5}{9\left( x-1 \right)}+\dfrac{31}{18\left( 4x+5 \right)}$ , to solve the above question first we have to bring all the functions to one side and only $f\left( x \right)$at one side. Then we have to take LCM and we should bring all the functions into one term that will be equal to the $f\left( x \right)$.

Complete step by step answer:
from the above question we have to find the $f\left( x \right)$ from
$\Rightarrow \dfrac{\left( 2x+3 \right)\left( 3x-4 \right)}{\left( x-1 \right)\left( 4x+5 \right)}=f\left( x \right)-\dfrac{5}{9\left( x-1 \right)}+\dfrac{31}{18\left( 4x+5 \right)}$
First, we will take the LCM of right-hand side terms except $f\left( x \right)$,
By this we will get, here the LCM is $18\left( x-1 \right)\left( 4x+5 \right)$, from this we will get,
$\Rightarrow -\dfrac{5}{9\left( x-1 \right)}+\dfrac{31}{18\left( 4x+5 \right)}=\dfrac{-10\left( 4x+5 \right)+31\left( x-1 \right)}{18\left( x-1 \right)\left( 4x+5 \right)}$
Now by replacing this in the above equation we will get,
$\Rightarrow \dfrac{\left( 2x+3 \right)\left( 3x-4 \right)}{\left( x-1 \right)\left( 4x+5 \right)}=f\left( x \right)+\dfrac{-10\left( 4x+5 \right)+31\left( x-1 \right)}{18\left( x-1 \right)\left( 4x+5 \right)}$
Now we have to bring all the terms except $f\left( x \right)$ from right hand side to the left-hand side of the equation,
By shifting the terms from left hand side to the right-hand side of the equation we will get,
$\Rightarrow f\left( x \right)=\dfrac{\left( 2x+3 \right)\left( 3x-4 \right)}{\left( x-1 \right)\left( 4x+5 \right)}-\dfrac{-10\left( 4x+5 \right)+31\left( x-1 \right)}{18\left( x-1 \right)\left( 4x+5 \right)}$
Now here the LCM of right-hand side of the equation is also $18\left( x-1 \right)\left( 4x+5 \right)$
from this we will get,
$\Rightarrow f\left( x \right)=\dfrac{18\left( 2x+3 \right)\left( 3x-4 \right)+10\left( 4x+5 \right)-31\left( x-1 \right)}{18\left( x-1 \right)\left( 4x+5 \right)}$
Now we have to multiply the terms in the numerator of right-hand side of the equation,
By multiplying we will get,
$\Rightarrow f\left( x \right)=\dfrac{108{{x}^{2}}+18x-216+40x+50-31x+31}{18\left( x-1 \right)\left( 4x+5 \right)}$
Now by further simplification we will get,
$\Rightarrow f\left( x \right)=\dfrac{108{{x}^{2}}+27x-135}{18\left( x-1 \right)\left( 4x+5 \right)}$
Now we will multiply the terms of denominator except $18$
By multiplying we will get,
$\Rightarrow f\left( x \right)=\dfrac{108{{x}^{2}}+27x-135}{18\left( 4{{x}^{2}}+x-5 \right)}$
Now we will take $27$common from all the terms in the numerator,
By taking common $27$from all the terms in the numerator we will get,
$\Rightarrow f\left( x \right)=\dfrac{27\left( 4{{x}^{2}}+x-5 \right)}{18\left( 4{{x}^{2}}+x-5 \right)}$
Now we will eliminate all the like terms in the numerator and denominator,
By eliminating the terms, we will get,
$\Rightarrow f\left( x \right)=\dfrac{27}{18}$
Now by further simplification we will get,
$\Rightarrow f\left( x \right)=\dfrac{3}{2}$

Therefore, the $f\left( x \right)=\dfrac{3}{2}$.

Note: Students should be very careful while doing the calculation part, if any sign is changed due to the negligence the whole answer will be wrong. Students should be careful while taking the LCM, if students write lcm as $162\left( x-1 \right)\left( 4x+5 \right)$instead of $18\left( x-1 \right)\left( 4x+5 \right)$ the whole answer will be wrong.