Question

How to find the factors of \[\left( {\dfrac{{x - 4}}{{x - 8}}} \right) + \left( {\dfrac{{x - 5}}{{x - 7}}} \right) = \dfrac{{x + 143}}{{{x^2} - 15x + 56}}\]?

Answer 1

Hint: This is a simple question of algebra which is based on the concept of LCM. We will first find the LCM of terms on the LHS of the equation and then equate it to the RHS. We will realise that the denominators on both sides of the equation would be the same, so we will be able to simplify easily and find the value of x.

Complete step by step answer:
Now let us consider the LHS:-
\[ \Rightarrow \]\[\left( {\dfrac{{x - 4}}{{x - 8}}} \right) + \left( {\dfrac{{x - 5}}{{x - 7}}} \right)\]
Now we will take LCM in both the denominators which means multiplying and dividing the \[\left( {\dfrac{{x - 4}}{{x - 8}}} \right)\]terms by (x-7) and the \[\left( {\dfrac{{x - 5}}{{x - 7}}} \right)\] by (x-8),
\[ \Rightarrow \]\[\left( {\dfrac{{x - 4}}{{x - 8}}} \right)\]\[\left( {\dfrac{{x - 7}}{{x - 7}}} \right)\]+\[\left( {\dfrac{{x - 5}}{{x - 7}}} \right)\]\[\left( {\dfrac{{x - 8}}{{x - 8}}} \right)\]
Now since denominators of both terms are equal we write it as,
\[ \Rightarrow \]\[\dfrac{{\left( {x - 4} \right)\left( {x - 7} \right) + \left( {x - 5} \right)\left( {x - 8} \right)}}{{\left( {x - 8} \right)\left( {x - 7} \right)}}\]
Now we will multiply the brackets in the numerator as well as the denominator,
\[ \Rightarrow \]\[\dfrac{{{x^2} - 11x + 28 + {x^2} - 13x + 40}}{{{x^2} - 15x + 56}}\]
Which can be further simplified as,
\[ \Rightarrow \]\[\dfrac{{2{x^2} - 24x + 68}}{{{x^2} - 15x + 56}}\]
Now we will equate LHS and RHS,
\[ \Rightarrow \]\[\dfrac{{2{x^2} - 24x + 68}}{{{x^2} - 15x + 56}}\]=\[\dfrac{{x + 143}}{{{x^2} - 15x + 56}}\]
Since the denominators in both LHS and RHS are equal, we can cancel it ,
Therefore,
\[ \Rightarrow \]\[2{x^2} - 24x + 68\]=\[x + 143\]
Now we will transfer all the terms in RHS to LHS,
\[ \Rightarrow \]\[2{x^2} - 24x + 68\] - (\[x + 143\])
Which can be further simplified as,
\[ \Rightarrow \]\[2{x^2} - 25x - 75\]=\[0\]
Now we will use the middle term formula to break the middle term of the quadratic equation,
\[ \Rightarrow \]\[2{x^2} - 30x + 5x - 75\]=\[0\] \[\xrightarrow{{}}eqnA\]
Let us consider,
\[ \Rightarrow \]\[2{x^2} - 30x\]
Now we will take \[2x\]common, giving us,
\[ \Rightarrow \]\[2x\left( {x - 15} \right)\] \[\xrightarrow{{}}eqn1\]
Let us consider,
\[ \Rightarrow \]\[5x - 75\]
Now we will take 5 common, giving us,
\[ \Rightarrow \] \[5\left( {x - 15} \right)\] \[\xrightarrow{{}}eqn2\]
Sub eqn 1 and 2 in eqn A,
\[ \Rightarrow \] \[2x\left( {x - 15} \right)\]+ \[5\left( {x - 15} \right)\]= \[0\]
Taking \[\left( {x - 15} \right)\]common,
\[ \Rightarrow \]\[\left( {x - 15} \right)\] \[\left( {2x + 5} \right)\]= \[0\]
Now, either
\[\left( {x - 15} \right)\]=\[0\] OR \[\left( {2x + 5} \right)\]=\[0\]
Therefore,
\[x\]=15, -\[\dfrac{5}{2}\] are the solutions.
An extraneous solution is a solution of a transformed equation but not the root of the original equation since they are excluded from the domain.
The points which will be excluded from domain are points at which the equation tends to infinity,
i.e. when the denominator tends to zero. This happens when,
\[ \Rightarrow \]\[{x^2} - 15x + 56\]=\[0\]
\[ \Rightarrow \]\[\left( {x - 8} \right)\left( {x - 7} \right)\]=\[0\]
Now, either
\[\left( {x - 8} \right)\]=\[0\]OR \[\left( {x - 7} \right)\]=\[0\]
Therefore,
\[x\]=7, 8 are the extraneous solutions of the equation.

Note: There are chances of making silly mistakes while taking LCM. Always keep in mind to be careful about the sign conventions during solving. Generally while solving such questions we forget to check whether the solutions that we obtained are extraneous or not. So always keep in mind to check for the extraneous solutions.