How to solve :- 3(3x-1/2x+3)-2(2x+3/3x-1)=5

Welcome to my article 3(3x-1/2x+3)-2(2x+3/3x-1)=5. 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 3(3x-1/2x+3)-2(2x+3/3x-1)=5, read and understand it carefully till the end.

Let us know how to solve this question 3(3x-1/2x+3)-2(2x+3/3x-1)=5.

First write the question on the page of the notebook.

3(3x-1/2x+3)-2(2x+3/3x-1)=5

By writing this question correctly in this way,

\displaystyle 3\left( {3x+\frac{{-1}}{2}x+3} \right)-2\left( {2x+\frac{3}{3}x-1} \right)=5

Let’s solve it

\displaystyle 3\left( {3x-\frac{1}{2}x+3} \right)-2\left( {2x+\frac{3}{3}x-1} \right)=5

\displaystyle 3\left( {3x-\frac{1}{2}x+3} \right)-2\left( {2x+1x-1} \right)=5

\displaystyle 3\left( {3x-\frac{x}{2}+3} \right)-2\left( {2x+1x-1} \right)=5

\displaystyle 3\left( {\frac{{2\times 3x-x}}{2}+3} \right)-2\left( {2x+1x-1} \right)=5

\displaystyle 3\left( {\frac{{6x-x}}{2}+3} \right)-2\left( {2x+1x-1} \right)=5

\displaystyle 3\left( {\frac{{5x}}{2}+3} \right)-2\left( {2x+1x-1} \right)=5

\displaystyle 3\left( {\frac{{5x+3\times 2}}{2}} \right)-2\left( {2x+1x-1} \right)=5

\displaystyle 3\left( {\frac{{5x+6}}{2}} \right)-2\left( {2x+1x-1} \right)=5

\displaystyle \frac{{3\times 5x+6\times 3}}{2}-2\left( {2x+1x-1} \right)=5

\displaystyle \frac{{15x+18}}{2}-2\left( {2x+1x-1} \right)=5

\displaystyle \frac{{15x+18}}{2}-2\left( {3x-1} \right)=5

\displaystyle \frac{{15x+18}}{2}-2\times 3x+2\times 1=5

\displaystyle \frac{{15x+18}}{2}-6x+2=5

\displaystyle \frac{{15x+18-2\times 6x+2\times 2}}{2}=5

\displaystyle \frac{{15x+18-12x+4}}{2}=5

\displaystyle \frac{{15x-12x+18+4}}{2}=5

\displaystyle \frac{{3x+22}}{2}=5

\displaystyle 3x+22=5\times 2

\displaystyle 3x+22=10

\displaystyle 3x=10-22

\displaystyle 3x=-12

\displaystyle x=-\frac{{12}}{3}

\displaystyle x=-4 Answer

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