# How to solve :- 2x/x-3+1/2x+3+3x+9/(x-3)(2x+3)=0

Welcome to my article 2x/x-3+1/2x+3+3x+9/(x-3)(2x+3)=0. 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 2x/x-3+1/2x+3+3x+9/(x-3)(2x+3)=0, read and understand it carefully till the end.

Let us know how to solve this question 2x/x-3+1/2x+3+3x+9/(x-3)(2x+3)=0.

First write the question on the page of the notebook.

## 2x/x-3+1/2x+3+3x+9/(x-3)(2x+3)=0

By writing this question correctly in this way,

## \displaystyle \frac{{2x}}{x}-3+\frac{1}{2}x+3+3x+\frac{9}{{\left( {x-3} \right)}}\times \left( {2x+3} \right)=0

Now let’s solve it like this,

We are seeing that there are equal numbers above and below.

Therefore, having the same number at the above and below, we remove it as follows,

\displaystyle \frac{2}{1}-3+\frac{1}{2}x+3+3x+\frac{9}{{\left( {x-3} \right)}}\times \left( {2x+3} \right)=0

\displaystyle \frac{2}{1}+\frac{1}{2}x+3x+\frac{9}{{\left( {x-3} \right)}}\times \left( {2x+3} \right)=0

\displaystyle \frac{2}{1}+\frac{1}{2}x+3x+\frac{{9\times \left( {2x+3} \right)}}{{\left( {x-3} \right)}}=0

\displaystyle \frac{2}{1}+\frac{1}{2}x+3x+\frac{{9\times 2x+9\times 3}}{{\left( {x-3} \right)}}=0

\displaystyle \frac{2}{1}+\frac{1}{2}x+3x+\frac{{18x+27}}{{\left( {x-3} \right)}}=0

\displaystyle \frac{2}{1}+\frac{x}{2}+\frac{{3x}}{1}+\frac{{18x+27}}{{\left( {x-3} \right)}}=0

\displaystyle \frac{2}{1}+\frac{x}{2}+\frac{{3x}}{1}+\frac{{18x+27}}{{\left( {x-3} \right)}}=0

\displaystyle \frac{{2\times 2+x+3x\times 2}}{2}+\frac{{18x+27}}{{\left( {x-3} \right)}}=0

\displaystyle \frac{{4+x+6x}}{2}+\frac{{18x+27}}{{\left( {x-3} \right)}}=0

\displaystyle \frac{{4+7x}}{2}+\frac{{18x+27}}{{\left( {x-3} \right)}}=0

\displaystyle \frac{{\left( {x-3} \right)\left( {4+7x} \right)+2\left( {18x+27} \right)}}{{2\left( {x-3} \right)}}=0

\displaystyle \frac{{4\left( {x-3} \right)+7x\left( {x-3} \right)+2\left( {18x} \right)+2\left( {27} \right)}}{{2\left( {x-3} \right)}}=0

\displaystyle \frac{{4\times x-4\times 3+7x\times x-7x\times 3+2\times 18x+2\times 27}}{{2\left( {x-3} \right)}}=0

\displaystyle \frac{{4x-12+7{{x}^{2}}-21x+36x+54}}{{2\left( {x-3} \right)}}=0

\displaystyle \frac{{7{{x}^{2}}+4x+36x-21x+54-12}}{{2\left( {x-3} \right)}}=0

\displaystyle \frac{{7{{x}^{2}}+40x-21x+54-12}}{{2\left( {x-3} \right)}}=0

\displaystyle \frac{{7{{x}^{2}}+19x+54-12}}{{2\left( {x-3} \right)}}=0

\displaystyle \frac{{7{{x}^{2}}+19x+42}}{{2\left( {x-3} \right)}}=0

\displaystyle \frac{{7{{x}^{2}}+19x+42}}{1}=0\times 2\left( {x-3} \right)

\displaystyle \frac{{7{{x}^{2}}+19x+42}}{1}=0

\displaystyle 7{{x}^{2}}+19x+42=0

\displaystyle x=\frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}

Once is standard form, identity a, b , and c from the original equation and plug them in to the quadratic formula :-

\displaystyle 7{{x}^{2}}+19x+42=0

a = 7

b = 19

c = 42

\displaystyle x=\frac{{-19\pm \sqrt{{{{{\left( {19} \right)}}^{2}}-4\times 42\times 7}}}}{{2\times 7}}

\displaystyle x=\frac{{-19\pm \sqrt{{361-28\times 42}}}}{{14}}

\displaystyle x=\frac{{-19\pm \sqrt{{361-1176}}}}{{14}}