Solve the following expression: $$2a-\left[ 4b-\left\{ 4a-\left( 3b-\overline{2a+2b} \right) \right\} \right] $$.

Hint: In this question it is given that we have to solve $$2a-\left[ 4b-\left\{ 4a-\left( 3b-\overline{2a+2b} \right) \right\} \right] $$.
So for this simplification we have to use the BODMAS rule which stands for Bracket, Of, Division, Multiplication, Addition, and Subtraction. According to BODMAS rule, the brackets have to be solved first followed by powers or roots (i.e. of), then Division, Multiplication, Addition and at the end Subtraction.
Complete step-by-step solution:
Given,
$$2a-\left[ 4b-\left\{ 4a-\left( 3b-\overline{2a+2b} \right) \right\} \right] $$
First of all we simply the quantities which is under first bracket,
Therefore,
$$2a-\left[ 4b-\left\{ 4a-\left( 3b-\overline{2a+2b} \right) \right\} \right] $$
=$$2a-\left[ 4b-\left\{ 4a-\left( 3b-2a-2b\right) \right\} \right] $$ [by omitting the overline]
=$$2a-\left[ 4b-\left\{ 4a-\left( 3b-2b-2a\right) \right\} \right] $$
=$$2a-\left[ 4b-\left\{ 4a-\left( b-2a\right) \right\} \right] $$
=$$2a-\left[ 4b-\left\{ 4a-b+2a\right\} \right] $$ [by omitting the first bracket]
=$$2a-\left[ 4b-\left\{ 4a+2a-b\right\} \right] $$
=$$2a-\left[ 4b-\left\{ 6a-b\right\} \right] $$
=$$2a-\left[ 4b-6a+b\right] $$ [by omitting the second bracket]
=$$2a-\left[ 4b+b-6a\right] $$
=$$2a-\left[ 5b-6a\right] $$
=$$2a-5b+6a$$
=$$2a+6a-5b$$
=$$8a-5b$$
Which is our required solution.
Note: While applying the BODMAS rule you have to keep in mind that you have to first solve the quantities under the first brackets ‘()’, after solving you have to go for second brackets ‘{}’ and then third brackets ‘[]’.