Solve the following equation by transposing the term \[2a - 3 = 5\]

Hint: In the given question we have to find the value of using a transposing method.
Before solving first, we must know what the transposing method is. So, the transposing method is the process of isolating across the equal sign of the equation.
The transposition method is one of the linear equation methods used to solve the linear equations.
We know that in complicated type equations, the two sides of an equation contain both variables and constants, in that case, we should simplify the equation in simple forms and transpose the terms that contain variables on both LHS and RHS.
In this question, keep variables at one side and constant terms on the other side of the ‘\[ = \]’ sign in the given equation.

Complete step-by-step answer:
 It is given that the equation is \[2a - 3 = 5\].
Now, put variables at one side and constant terms on the other side of the\[ = \] sign in the given equation.
Also, when \[ - 3\] goes to another side of ‘\[ = \]’ sign it will be converted to \[ + 3\].
\[2a = 5 + 3\]
Now simplify and find the value of the variable \[a\],
\[2a = 8\]
Now transform \[2\]to the other side of the ‘\[ = \]’ sign (RHS), after transforming \[2\], it will divide \[8\].
\[a = \dfrac{8}{2}\]
\[a = 4\]
Hence, we solved the equation by transposing the method and the value of \[a = 4\] the above equation.
To verify the result, put the value \[a = 4\] obtained on solving into the given equation \[2a - 3 = 5\].We get,
\[2 \times 4 - 3 = 5\]
\[5 = 5\]
\[LHS = RHS\].
Hence verified.

Note: Whenever you take a term from one side of ‘\[ = \]’sign to another side, always keep in mind to revert the sign of the term. for example: - convert ‘\[ + \]’ into ‘\[ - \]’and ‘\[ - \]’into ‘\[ + \]’, ‘\[ \times \]’into \[ \div \] and \[ \div \] into \[ \times \]
Always calculate the terms properly.
Remember to isolate variable and constant terms means we have to introduce terms other than ‘\[a\]’ on one side and all numbers on the other side.