Welcome to my article a=x(2x+3)-4(x+1)-2x(x-1/2) ?This question is taken from the simplification lesson.

The solution of this question has been explained in a very simple way by a well-known teacher by doing addition, subtraction, and fractions.

For complete information on how to solve this question a=x(2x+3)-4(x+1)-2x(x-1/2)? read and understand it carefully till the end.

Let us know how to solve this question a=x(2x+3)-4(x+1)-2x(x-1/2)?

First write the question on the page of the notebook.

## a=x(2x+3)-4(x+1)-2x(x-1/2)

writing this question correctly in this way,

** \displaystyle a=x\left( {2x+3} \right)-4\left( {x+1} \right)-2x\left( {x+\frac{{-1}}{2}} \right) ,**

**Let us now solve this line by line**.

\displaystyle a=x\left( {2x+3} \right)-4\left( {x+1} \right)-2x\left( {x-\frac{1}{2}} \right) ,

\displaystyle a=x\left( {2x+3} \right)-4\left( {x+1} \right)-2x\times x-2x\times -\frac{1}{2} ,

\displaystyle a=x\left( {2x+3} \right)-4\left( {x+1} \right)-2{{x}^{2}}+\frac{{2x}}{2} ,

\displaystyle a=x\left( {2x+3} \right)-4\left( {x+1} \right)-2{{x}^{2}}+\frac{2}{2}x ,

\displaystyle a=x\left( {2x+3} \right)-4\left( {x+1} \right)-2{{x}^{2}}+x ,

\displaystyle a=x\left( {2x+3} \right)-4\times x-4\times 1-2{{x}^{2}}+x ,

\displaystyle a=x\left( {2x+3} \right)-4x-4-2{{x}^{2}}+x ,

\displaystyle a=x\left( {2x+3} \right)-4x+x-2{{x}^{2}}-4 ,

\displaystyle a=x\left( {2x+3} \right)-3x-2{{x}^{2}}-4 ,

\displaystyle a=x\times 2x+x\times 3-3x-2{{x}^{2}}-4 ,

\displaystyle a=2{{x}^{2}}+3x-3x-2{{x}^{2}}-4 ,

\displaystyle a=2{{x}^{2}}-2{{x}^{2}}+3x-3x-4 ,

\displaystyle a=3x-3x-4 ,

\displaystyle a=-4 ,

In this way, after doing this question line by line, its intended solution is –

**a = -4 [answer]**

This article** How to solve:- a=x(2x+3)-4(x+1)-2x(x-1/2) ?** has been completely solved by tireless effort from our side, still if any error remains in it then definitely write us your opinion in the comment box. If you like or understand the methods of solving all the questions in this article, then send it to your friends who are in need.

Note: If you have any such question, then definitely send it by writing in our comment box to get the answer.

Your question will be answered from our side.

Thank you once again from our side for reading or understanding this article completely.