Question

How do you solve the expression $\dfrac{{x - 2}}{{x - 3}} + \dfrac{{x - 3}}{{x - 2}} = \dfrac{{2{x^2}}}{{{x^2} - 5x + 6}}$?

Answer 1

Hint: We will first of all take the least common multiple on the left hand side and then we will observe that the left hand right hand side has the same denominator and the rest is a quadratic equation.

Complete step-by-step solution:
We are given that we are required to solve $\dfrac{{x - 2}}{{x - 3}} + \dfrac{{x - 3}}{{x - 2}} = \dfrac{{2{x^2}}}{{{x^2} - 5x + 6}}$.
Taking the least common multiple of the denominators on the left hand side, we will then obtain the following equation with us:-
$ \Rightarrow \dfrac{{{{\left( {x - 2} \right)}^2} + {{\left( {x - 3} \right)}^2}}}{{\left( {x - 3} \right)\left( {x - 2} \right)}} = \dfrac{{2{x^2}}}{{{x^2} - 5x + 6}}$
We can write the denominator of the right hand side as follows:-
$ \Rightarrow \dfrac{{{{\left( {x - 2} \right)}^2} + {{\left( {x - 3} \right)}^2}}}{{\left( {x - 3} \right)\left( {x - 2} \right)}} = \dfrac{{2{x^2}}}{{\left( {x - 3} \right)\left( {x - 2} \right)}}$ ……………………(1)
Since, we know that we have a formula given by the following expression with us:-
$ \Rightarrow {\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$
Replacing a by x and b by 2 in the above mentioned expression, we will then obtain the following expression:-
$ \Rightarrow {\left( {x - 2} \right)^2} = {x^2} + {2^2} - 2 \times 2 \times x$
Simplifying the right hand side of the above equation, we will then obtain the following equation with us:-
$ \Rightarrow {\left( {x - 2} \right)^2} = {x^2} + 4 - 4x$ ……………..(2)
Replacing a by x and b by 3 in the above mentioned expression, we will then obtain the following expression:-
$ \Rightarrow {\left( {x - 3} \right)^2} = {x^2} + {3^2} - 2 \times 3 \times x$
Simplifying the right hand side of the above equation, we will then obtain the following equation with us:-
$ \Rightarrow {\left( {x - 3} \right)^2} = {x^2} + 9 - 6x$ ……………..(3)
Putting the equation number 2 and equation number 3 in equation number 1, we will then obtain the following expression with us:-
$ \Rightarrow \dfrac{{{x^2} + 4 - 4x + {x^2} + 9 - 6x}}{{\left( {x - 3} \right)\left( {x - 2} \right)}} = \dfrac{{2{x^2}}}{{\left( {x - 3} \right)\left( {x - 2} \right)}}$
Simplifying the numerator in the left hand side, we will then obtain the following expression:-
$ \Rightarrow \dfrac{{2{x^2} + 13 - 10x}}{{\left( {x - 3} \right)\left( {x - 2} \right)}} = \dfrac{{2{x^2}}}{{\left( {x - 3} \right)\left( {x - 2} \right)}}$
Cross – multiplying the above equation to obtain the following expression with us:-
$ \Rightarrow \left( {x - 3} \right)\left( {x - 2} \right)\left( {2{x^2} + 13 - 10x} \right) = 2{x^2}\left( {x - 3} \right)\left( {x - 2} \right)$
Taking all the terms from the right hand side of above expression to left hand side, we will then obtain the following expression with us:-
$ \Rightarrow \left( {x - 3} \right)\left( {x - 2} \right)\left( {2{x^2} + 13 - 10x} \right) - 2{x^2}\left( {x - 3} \right)\left( {x - 2} \right) = 0$
Taking $\left( {x - 3} \right)\left( {x - 2} \right)$ common from the left hand side of the above equation, we will then obtain the following expression with us:-
$ \Rightarrow \left( {x - 3} \right)\left( {x - 2} \right)\left\{ {\left( {2{x^2} + 13 - 10x} \right) - 2{x^2}} \right\} = 0$
Simplifying the latter part, we have:-
$ \Rightarrow \left( {x - 3} \right)\left( {x - 2} \right)\left( {13 - 10x} \right) = 0$
Thus, we have x = 3, 2 or $\dfrac{{13}}{{10}}$.

Note: The students must notice that we did not cancel out the denominators of the left hand side and right hand side, even when they were equal. That is because we were not given in the question stem that x cannot be equal to 3 or 2 because if x = 2 or x = 3, then we can never cancel out 0 from both sides and it would have become not defined. Therefore, we took the possibility of x = 2 and x = 3 as well.