Solve $\dfrac{{6x}}{{x + 4}} + 4 = \dfrac{{2x + 2}}{{x - 1}}$?

Hint: Here first simplify the equation by taking L.C.M, and by cross multiplying the denominators we will get a quadratic equation of the form $a{x^2} + bx + c = 0$ which we will solve by using quadratic formula which is given by $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$, where a, b, and c are real numbers, now substituting the values in the formula we will get the required result.

Complete step by step answer:
Given equation is $\dfrac{{6x}}{{x + 4}} + 4 = \dfrac{{2x + 2}}{{x - 1}}$,
First take L.C.M on the left hand side of the equation, we get,
$ \Rightarrow \dfrac{{6x + 4x + 16}}{{x + 4}} = \dfrac{{2x + 2}}{{x - 1}}$,
Now simplifying we get,
$ \Rightarrow \dfrac{{10x + 16}}{{x + 4}} = \dfrac{{2x + 2}}{{x - 1}}$,
Now taking out the common terms in the numerator and eliminating them we get,
$ \Rightarrow \dfrac{{2\left( {5x + 8} \right)}}{{x + 4}} = \dfrac{{2\left( {x + 1} \right)}}{{x - 1}}$,
Now simplifying we get,
$ \Rightarrow \dfrac{{5x + 8}}{{x + 4}} = \dfrac{{x + 1}}{{x - 1}}$
Now cross multiply the denominators, we get,
$ \Rightarrow \left( {5x + 8} \right)\left( {x - 1} \right) = \left( {x + 1} \right)\left( {x + 4} \right)$,
Now multiplying we get,
$ \Rightarrow 5{x^2} - 5x + 8x - 8 = {x^2} + 4x + x + 4$,
Again simplifying we get,
$ \Rightarrow 5{x^2} + 3x - 8 = {x^2} + 5x + 4$,
Now taking all terms to one side we get,
$ \Rightarrow 5{x^2} + 3x - 8 - {x^2} - 5x - 4$,
Now simplifying we get,
$ \Rightarrow 4{x^2} - 2x - 12 = 0$,
Now taking out the common terms we get,
$ \Rightarrow 2\left( {2{x^2} - x - 6} \right) = 0$,
Now simplifying we get,
$ \Rightarrow 2{x^2} - x - 6 = 0$,
Now the equation is a quadratic equation of the form $a{x^2} + bx + c = 0$ which we will solve by using quadratic formula which is given by $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$, where a, b, and c are real numbers,
Here $a = 2$, $b = - 1$ and $c = - 6$,
Now substituting the values in the formula we get,
$ \Rightarrow x = \dfrac{{ - \left( { - 1} \right) \pm \sqrt {{{\left( { - 1} \right)}^2} - 4\left( 2 \right)\left( { - 6} \right)} }}{{2\left( 2 \right)}}$,
Now simplifying we get,
$ \Rightarrow x = \dfrac{{1 \pm \sqrt {1 - \left( { - 48} \right)} }}{4}$,
Now simplifying we get,
$ \Rightarrow x = \dfrac{{1 \pm \sqrt {1 + 48} }}{4}$,
Now again simplifying we get,
$ \Rightarrow x = \dfrac{{1 \pm \sqrt {49} }}{4}$,
Now taking out the square root we get,
$ \Rightarrow x = \dfrac{{1 \pm 7}}{4}$,
Further simplifying we get,
$ \Rightarrow x = \dfrac{{1 + 7}}{4}$and $x = \dfrac{{1 - 7}}{4}$,
Now simplifying we get,
$ \Rightarrow x = \dfrac{8}{4}$and $x = \dfrac{{ - 6}}{4}$,
Further simplifying we get,
$ \Rightarrow x = 2$and $x = \dfrac{{ - 3}}{2}$.
So, when the equation is solved we get $x = 2,\dfrac{{ - 3}}{2}$.

$\therefore $ The value of $x$ when the given equation $\dfrac{{6x}}{{x + 4}} + 4 = \dfrac{{2x + 2}}{{x - 1}}$ is solved will be equal to $2$ and $\dfrac{{ - 3}}{2}$.

Note: Quadratic equation formula is a method of solving quadratic equations, but we should keep in mind that we can also solve the equation using completely the square, and we can cross check the values of $x$ by using the above formula. Also we should always convert the coefficient of${x^2} = 1$, to easily solve the equation by this method, and there are other methods to solve such kind of solutions, other method used to solve the quadratic equation is by factoring method, in this method we should obtain the solution factorising quadratic equation terms. In these types of questions, we can solve by using quadratic formula i.e., $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.