Question

Solve each of the following equations:
$\dfrac{{2x}}{{x - 3}} + \dfrac{1}{{2x + 3}} + \dfrac{{3x + 9}}{{(x - 3)(2x + 3)}} = 0\,\,;\,x \ne 3,\,\,\,x \ne \dfrac{{ - 3}}{2}$

Answer 1

Hint: If it is given in the question that we can’t take a particular value of x, then neglect that value. If a polynomial equation is forming then take only those values of x which are not neglected in the question statement. We can use the factorization method in this question to find multiple values of x if required.

Complete step by step answer:
We have,
$\dfrac{{2x}}{{x - 3}} + \dfrac{1}{{2x + 3}} + \dfrac{{3x + 9}}{{(x - 3)(2x + 3)}} = 0$
Taking L.C.M of the equation,
$\dfrac{{(2x)(2x + 3) + (x - 3) + (3x + 9)}}{{(x - 3)(2x + 3)}} = 0$
Multiplying $(x - 3)(2x + 3)$ with $0$ and we get $0$ after multiplication.
So,
$(2x)(2x + 3) + (x - 3) + (3x + 9) = 0$
$4{x^2} + 6x + x - 3 + 3x + 9 = 0$
On adding, we get
$4{x^2} + 10x + 6 = 0$
We can also write $10x$ as $6x + 4x$.
$4{x^2} + 6x + 4x + 6 = 0$
Now,
$(4{x^2} + 6x) + (4x + 6) = 0$
Taking out common $2x$ from $4{x^2} + 6x$ and $2$ from $4x + 6$.
Therefore,
$2x(2x + 3) + 2(2x + 3) = 0$
Taking common $(2x + 3)$ from the equation.
$(2x + 3)(2x + 2) = 0$
Now, there are two cases: either $2x + 3$ can be zero or $2x + 2$ can be zero.
Case I:
$2x + 3 = 0$
$2x = - 3$
$x = \dfrac{{ - 3}}{2}$
But it is given in the question that x cannot be equal to $ - \dfrac{3}{2}$.
So, we cannot take this value.
Case II:
$2x + 2 = 0$
$2x = - 2$
$x = - 1$
There is no restriction for taking the value $x = - 1$.
So, we can take this value.
Hence, the value of x is $ - 1$.

Note:
We have to be careful while solving a question whenever there is some extra statement or restriction. If we ignore that, then the answer might come wrong. Here in this question if we neglect the restriction then there would be two answers which can be true because the equation which we formed is a quadratic one. But we have only one answer to this question.