Solve the following equation: \[\dfrac{{\left( {Y - 1} \right)}}{3} - \dfrac{{\left( {Y - 2} \right)}}{4} = 1\].
A. None of these
B. \[ - 10\]
C. \[10\]
D. Does not exist

Hint: Here, we will have to first make the denominator (in the left hand side of the equation) equal algebraically i.e. by multiplying with common divisible of both the terms present in the denominator i.e. ‘\[12\]’. As a result, solving the equation periodically using certain mathematical concepts like properties/laws, BODMAS, etc. the significant value is obtained.

Complete step-by-step answer:
Since, we have given that
\[\dfrac{{\left( {Y - 1} \right)}}{3} - \dfrac{{\left( {Y - 2} \right)}}{4} = 1\]
As a result, we have to find/solve value of the given equation ‘\[\dfrac{{\left( {Y - 1} \right)}}{3} - \dfrac{{\left( {Y - 2} \right)}}{4} = 1\] ‘,
Hence, solving the given equation algebraically the given equation can be solved
Therefore, solving the respective equation, we get
That is multiplying the whole equation by ‘\[12\]’, we get
 \[ \Rightarrow 12 \times \dfrac{{\left( {Y - 1} \right)}}{3} - 12 \times \dfrac{{\left( {Y - 2} \right)}}{4} = 1 \times 12\]
As a result, cancelling (or dividing) the certain terms in left hand side, we get
\[ \Rightarrow 4 \times \left( {Y - 1} \right) - 3 \times \left( {Y - 2} \right) = 12\]
Now, since multiplying the bracket by the respective term present outside the bracket that is also known as ‘distributive law or the property which seems to be \[a \times \left( {b + c} \right) = ab + ac\]’, we get
\[ \Rightarrow 4Y - 4 - 3Y + 6 = 12\]
Now,
Simplifying the equation (i.e. by adding or subtracting) with the equal variable and constants, we get
\[ \Rightarrow Y + 2 = 12\]
\[ \Rightarrow Y = 12 - 2\]
Hence, we get
\[ \Rightarrow Y = 10\]
So, the correct answer is “Option C”.

Note: Whenever we come up with this type of problem, always remember that the denominator must be same to add or subtract the respective equation where, for multiplication or division it does not need to make the same Also, while doing such calculations just try to mug up the algebraic property such as associative law, commutative law, distributive law, etc. Like in this question, here we used the distributive law to multiply ‘\[3,4\]’ to its respective bracket.