Solve: $\dfrac{{15\left( {2 - y} \right) - 5\left( {y + 6} \right)}}{{1 - 3y}} = 10$

Hint:The value of x in $\dfrac{{15\left( {2 - y} \right) - 5\left( {y + 6} \right)}}{{1 - 3y}} = 10$ can be found by using the method of transposition. We will first cross multiply the terms of the equation and simplify the equation to find the value of x. Method of transposition involves doing the exact same mathematical thing on both sides of an equation with the aim of simplification in mind. This method can be used to solve various algebraic equations like the one given in question with ease.

Complete step by step answer:
We would use the method of transposition to find the value of x in $\dfrac{{15\left( {2 - y} \right) - 5\left( {y + 6} \right)}}{{1 - 3y}} = 10$.
Method of transposition involves doing the exact same thing on both sides of an equation with the aim of bringing like terms together and isolating the variable or the unknown term in order to simplify the equation and finding the value of the required parameter.Now, In order to find the value of x, we need to isolate x from the rest of the parameters such as constant terms.
So, $\dfrac{{15\left( {2 - y} \right) - 5\left( {y + 6} \right)}}{{1 - 3y}} = 10$
Cross multiplying the terms, we get,
$ \Rightarrow $$15\left( {2 - y} \right) - 5\left( {y + 6} \right) = 10\left( {1 - 3y} \right)$
Opening the brackets, we get,
$ \Rightarrow $$30 - 15y - 5y - 30 = 10 - 30y$

Now, isolating the terms consisting x by shifting the terms in the equation. So, we shift all the constant terms to right and terms consisting of variables to the left.
$ \Rightarrow $$ - 15y - 5y + 30y = 10 + 30 - 30$
Cancelling the terms with same magnitude and opposite signs, we get,
$ \Rightarrow $$ - 15y - 5y + 30y = 10$
Adding up the like terms,
$ \Rightarrow $$10y = 10$
Now, dividing both sides by ten, we get,
\[ \Rightarrow y = \dfrac{{10}}{{10}}\]
Cancelling common terms in numerator and denominator, we get,
\[ \Rightarrow y = 1\]

Hence, the value of y in $\dfrac{{15\left( {2 - y} \right) - 5\left( {y + 6} \right)}}{{1 - 3y}} = 10$ is $y = 1$.

Note: If we add, subtract, multiply or divide by the same number on both sides of a given algebraic equation, then both sides will remain equal. We must remember to reverse the signs of the terms while shifting the terms from one side of the equation to the other side. There is no fixed way of solving a given algebraic equation. Algebraic equations can be solved in various ways. Linear equations in one variable can be solved by a transposition method with ease.