Question

Write down preferably by inspection, the solution of the equation $\dfrac{\left( x-ab \right)}{\left( a+b \right)}+\dfrac{\left( x-ac \right)}{\left( a+c \right)}+\dfrac{\left( x-cb \right)}{\left( c+b \right)}=a+b+c$
(a) $ab+bc+ca$
(b) $ab-bc+ca$
(c) $ab+bc-ca$
(d) None of these

Answer 1

Hint: In order to solve the above equation we will use Hit and Trial method by inspection.
Substituting the value of $x$ in the above equation from the options and if it satisfies the given equation, then it is a required solution of the given equation.

Complete step-by-step answer:
We have equation given, $\dfrac{\left( x-ab \right)}{\left( a+b \right)}+\dfrac{\left( x-ac \right)}{\left( a+c \right)}+\dfrac{\left( x-cb \right)}{\left( c+b \right)}=a+b+c$
Now first we will go with first option,
 (a) $ab+bc+ca$
Putting, $x=ab+bc+ca$ in above equation L.H.S, we have
L.H.S
$=\dfrac{\left( ab+bc+ca-ab \right)}{\left( a+b \right)}+\dfrac{\left( ab+bc+ca-ac \right)}{\left( a+c \right)}+\dfrac{\left( ab+bc+ca-cb \right)}{\left( c+b \right)}$
After simplifying numerator in the above equation, we have
$=\dfrac{\left( bc+ca \right)}{\left( a+b \right)}+\dfrac{\left( ab+bc \right)}{\left( a+c \right)}+\dfrac{\left( ab+ca \right)}{(b+c)}$
Taking c common from first term, b common from second term and a common from third term we have,
$=\dfrac{c\left( a+b \right)}{\left( a+b \right)}+\dfrac{b\left( a+c \right)}{\left( a+c \right)}+\dfrac{a\left( b+c \right)}{\left( b+c \right)}$
We have some common element in the numerator and denominator of each term, so cancelling out the common element from numerator and denominator we have,
$=c+b+a$
Again after rearranging the terms we have,
$\begin{align}
  & =a+b+c \\
 & =\text{R}\text{.}H.S \\
\end{align}$
Hence, from this we conclude that $x=ab+bc+ca$ is the solution of the given equation (i) Similarly, we can check for option (b) and option (c), but since option (a) satisfies the given equation, no need to check for the other options and Hence,

So, the correct answer is “Option a”.

Note: Always remember that in order solve any complex equation, never go ahead with general process in M.C.Q type questions, instead try to attack the question by substituting the options provided in given question and if any option satisfies the the given equation, then that is the required answer