Solve for \[x\]: \[2(\dfrac{{2x - 1}}{{x + 3}}) - 3(\dfrac{{x + 3}}{{2x - 1}}) = 5{\kern 1pt} {\kern 1pt} ;x \ne - 3,\dfrac{1}{2}\]

Hint: It is advisable that we must know the basic properties of multiplication, division, subtraction and addition of algebraic equations with unknown variables. It is very important that we know how to check where the equation is not defined and we reject those values from the equation.

Complete step-by-step solution:
We are given with the equation\[2(\dfrac{{2x - 1}}{{x + 3}}) - 3(\dfrac{{x + 3}}{{2x - 1}}) = 5{\kern 1pt} {\kern 1pt} \],
Let us start with the left hand side of the equation and simplify it to smaller terms,
Let us multiply the constants outside the brackets inside and write the equation,
\[(\dfrac{{4x - 2}}{{x + 3}}) - (\dfrac{{3x + 9}}{{2x - 1}})\]
Now let us take the LCM of the denominators and write the equation accordingly,
\[\dfrac{{(2x - 1)(4x - 2) - (x + 3)(3x + 9)}}{{(x + 3)(2x - 1)}}\]
Now let us simplify our obtained numerator and write it in simplest form,
\[ = \dfrac{{(8{x^2} - 4x - 4x + 2) - (3{x^2} + 9x + 9x + 27)}}{{(x + 3)(2x - 1)}}\]
\[ = \dfrac{{8{x^2} - 4x - 4x + 2 - 3{x^2} - 9x - 9x - 27}}{{(x + 3)(2x - 1)}}\]
\[ = \dfrac{{5{x^2} - 26x - 25}}{{(x + 3)(2x - 1)}}\]
And this is equal to our given Right hand side of the equation i.e.
\[ \Rightarrow \dfrac{{5{x^2} - 26x - 25}}{{(x + 3)(2x - 1)}} = 5\]
\[ \Rightarrow 5{x^2} - 26x - 25 = 5{\kern 1pt} {\kern 1pt} (x + 3)(2x - 1)\]
After multiplying and simplifying our right hand side we get,
\[ \Rightarrow 5{x^2} - 26x - 25 = 5{\kern 1pt} {\kern 1pt} (2{x^2} - x + 6x - 3)\]
\[ \Rightarrow 5{x^2} - 26x - 25 = 10{x^2} + 25x - 15\]
Taking the terms of left hand side of our equation to the right hand side we get,
\[ \Rightarrow 5{x^2} + 51x + 10 = 0\]
\[ \Rightarrow 5{x^2} + 50x + x + 10 = 0\]
\[ \Rightarrow 5x(x + 10) + 1(x + 10) = 0\]
\[ \Rightarrow (5x + 1)(x + 10) = 0\]
This gives us that each of the factors is equal to zero i.e.
\[(5x + 1) = 0\],\[(x + 10) = 0\]
\[ \Rightarrow x = \dfrac{{ - 1}}{5}, - 10\]
Therefore our required solutions for the given equation is \[x = \dfrac{{ - 1}}{5}, - 10\]
Additional information: There are several types of equations: polynomial equations, trigonometric equations, logarithmic equations, exponential equations, differential equations and then a mixture of equations consisting of each of them. These help us to represent many real life examples in terms of mathematical expressions and solve accordingly as per our requirements.

Note: It is important that we note that the solutions or roots of the given equation cannot be \[ - 3,\dfrac{1}{2}\]as we notice that the value of the equation at the given values\[ - 3,\dfrac{1}{2}\]make the denominator not defined and we get an absurd result.