Question

Solve the given equation $\dfrac{{41x - 12}}{{{x^2} - 16}} = \dfrac{{4x + 3}}{{x - 4}}$

Answer 1

Hint: Here, we will first rewrite the denominator such that it is in the form of some algebraic identity. We will then simplify it using suitable identity. We will solve the equation further to get a quadratic equation. Then we will factorize the equation to get the required answer.

Formula Used:
${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$

Complete step-by-step answer:
First, we will rewrite the denominator as:
$\dfrac{{41x - 12}}{{{{\left( x \right)}^2} - {{\left( 4 \right)}^2}}} = \dfrac{{4x + 3}}{{x - 4}}$
Now, we will use the identity ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$ in the denominator of the LHS,
$ \Rightarrow \dfrac{{41x - 12}}{{\left( {x - 4} \right)\left( {x + 4} \right)}} = \dfrac{{4x + 3}}{{x - 4}}$
Now cancelling out same terms from both the terms, we get
$ \Rightarrow \dfrac{{41x - 12}}{{\left( {x + 4} \right)}} = \dfrac{{4x + 3}}{1}$
On cross multiplication, we get
$ \Rightarrow \left( {41x - 12} \right) = \left( {4x + 3} \right)\left( {x + 4} \right)$
Not multiplying the terms, we get
$ \Rightarrow \left( {41x - 12} \right) = 4x\left( {x + 4} \right) + 3\left( {x + 4} \right)$
Again multiplying using the distributive property, we get
$ \Rightarrow \left( {41x - 12} \right) = 4{x^2} + 16x + 3x + 12$
Rewriting the equation, we get
$ \Rightarrow 4{x^2} + 19x - 41x + 12 + 12 = 0$
Adding and subtracting like terms, we get
$ \Rightarrow 4{x^2} - 22x + 24 = 0$
Dividing both sides by 2, we get
$ \Rightarrow 2{x^2} - 11x + 12 = 0$
The above obtained equation is a quadratic equation, so we will factorize this equation to find the required solution.
Now, doing middle term split, we get
$ \Rightarrow 2{x^2} - 8x - 3x + 12 = 0$
Factoring out common terms, we get
$ \Rightarrow \left( {2x - 3} \right)\left( {x - 4} \right) = 0$
Now using zero product property, we get
$ \Rightarrow \left( {2x - 3} \right) = 0$
$ \Rightarrow x = \dfrac{3}{2}$
And,
$ \Rightarrow \left( {x - 4} \right) = 0$
$ \Rightarrow x = 4$
But, $x \ne 4$ because this will make the denominator 0 and hence, it will not be defined.
Thus, rejecting this value, we get the required value of $x$ as $\dfrac{3}{2}$.
Thus, this is the required answer.

Note: We can also check whether our answer is correct or not by substituting the value of $x$ in the given question.
Thus, it is given that:
$\dfrac{{41x - 12}}{{{x^2} - 16}} = \dfrac{{4x + 3}}{{x - 4}}$
Here, substituting $x = \dfrac{3}{2}$, we get,
LHS $ = \dfrac{{41\left( {\dfrac{3}{2}} \right) - 12}}{{{{\left( {\dfrac{3}{2}} \right)}^2} - 16}} = \dfrac{{\left( {\dfrac{{123 - 24}}{2}} \right)}}{{\dfrac{9}{4} - 16}}$
$ \Rightarrow $ LHS $ = \dfrac{{\left( {\dfrac{{123 - 24}}{2}} \right)}}{{\left( {\dfrac{{9 - 64}}{4}} \right)}} = \dfrac{{99 \times 2}}{{ - 55}} = \dfrac{{ - 9 \times 2}}{5} = \dfrac{{ - 18}}{5}$
Now,
RHS \[ = \dfrac{{4\left( {\dfrac{3}{2}} \right) + 3}}{{\dfrac{3}{2} - 4}} = \dfrac{{\left( {\dfrac{{12 + 6}}{2}} \right)}}{{\dfrac{{3 - 8}}{2}}} = \dfrac{{ - 18}}{5}\]
Clearly, LHS $ = $ RHS
Therefore, our answer is correct.
Hence, the required value of $x$ is $\dfrac{3}{2}$