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\[\dfrac{1}{{\left( {x - 1} \right)\left( {x - 2} \right)}} + \dfrac{1}{{\left( {x - 2} \right)\left( {x - 3} \right)}} = \dfrac{2}{3}\] where \[x \ne 1,x \ne 2,x \ne 3\]

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Last updated date: 22nd Jul 2024
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Answer
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Hint: While solving the equation make sure to remember \[x \ne 1,x \ne 2,x \ne 3\]
excluding the above values x can take any values as x =1 or x= 2,3 will make denominator zero.

Complete step-by-step answer:
Given that,
\[\dfrac{1}{{\left( {x - 1} \right)\left( {x - 2} \right)}} + \dfrac{1}{{\left( {x - 2} \right)\left( {x - 3} \right)}} = \dfrac{2}{3}\]
We will first find the LCM of the terms on LHS, for that we will multiply numerator and denominator of the first fraction by \[\left( {x - 3} \right)\] and that of second by \[\left( {x - 1} \right)\].
\[ \Rightarrow \dfrac{{\left( {x - 3} \right)}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}} + \dfrac{{\left( {x - 1} \right)}}{{\left( {x - 2} \right)\left( {x - 3} \right)\left( {x - 1} \right)}} = \dfrac{2}{3}\]
Now since the denominators are same we can add the numerators directly,
\[ \Rightarrow \dfrac{{\left( {x - 3} \right) + \left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}} = \dfrac{2}{3}\]
Simplifying the terms in the numerator,
\[ \Rightarrow \dfrac{{x - 3 + x - 1}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}} = \dfrac{2}{3}\]
On adding the terms in the numerator,
\[ \Rightarrow \dfrac{{2x - 4}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}} = \dfrac{2}{3}\]
Taking two common from the terms in the numerator of the LHS,
\[ \Rightarrow \dfrac{{2\left( {x - 2} \right)}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}} = \dfrac{2}{3}\]
Cancelling 2 from both the sides,
\[ \Rightarrow \dfrac{{\left( {x - 2} \right)}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}} = \dfrac{1}{3}\]
On cross multiplying we get,
\[ \Rightarrow 3\left( {x - 2} \right) = \left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)\]
Now we will solve the RHS separately,
\[ \Rightarrow \left[ {\left( {x - 1} \right)\left( {x - 2} \right)} \right]\left( {x - 3} \right)\]
Now on multiplying the first two brackets,
\[ \Rightarrow \left[ {x\left( {x - 2} \right) - 1\left( {x - 2} \right)} \right]\left( {x - 3} \right)\]
Now multiplying the terms inside the bracket,
\[ \Rightarrow \left[ {{x^2} - 2x - x + 2} \right]\left( {x - 3} \right)\]
On adding the same terms,
\[ \Rightarrow \left[ {{x^2} - 3x + 2} \right]\left( {x - 3} \right)\]
Now multiply the outside bracket with the terms inside the bracket,
\[ \Rightarrow x\left( {{x^2} - 3x + 2} \right) - 3\left( {{x^2} - 3x + 2} \right)\]
On multiplying we get,
\[ \Rightarrow {x^3} - 3{x^2} + 2x - 3{x^2} + 9x - 6\]
On adding the same terms,
\[ \Rightarrow {x^3} - 6{x^2} + 11x - 6\]
This is the simplified product of the three brackets,
\[ \Rightarrow 3\left( {x - 2} \right) = {x^3} - 6{x^2} + 11x - 6\]
This is to be further simplified,
\[ \Rightarrow 3x - 6 = {x^3} - 6{x^2} + 11x - 6\]
Taking all the terms on one side,
\[ \Rightarrow {x^3} - 6{x^2} + 11x - 3x - 6 + 6 = 0\]
On solving we get,
\[ \Rightarrow {x^3} - 6{x^2} + 8x = 0\]
Dividing both sides by x we get,
\[ \Rightarrow {x^2} - 6x + 8 = 0\]
This is quadratic equation, we will solve this factorisation method,
\[ \Rightarrow {x^2} - 4x - 2x + 8 = 0\]
Taking x common form first two terms and 2 common from last two terms,
\[ \Rightarrow x\left( {x - 4} \right) - 2\left( {x - 4} \right) = 0\]
Taking the brackets separately,
\[ \Rightarrow \left( {x - 4} \right)\left( {x - 2} \right) = 0\]
Thus the value of x can be 4 or 2. But in the question they have mentioned it cannot be 2.
So the final answer is that the value of x is 4.
So, the correct answer is “x = 4”.

Note: Note that, the simplification of the fractions is the most important step. Also note that the quadratic equation can also be solved by quadratic formula. But here the factorization is used because it was not difficult to factorize the numbers. The factors should be such that their product should be the last term and sum should be the middle term.