Question

Given \[\dfrac{1}{{x + 1}} + \dfrac{3}{{5x + 1}} = \dfrac{5}{{x + 4}}\]. Solve for $x$.

Answer 1

Hint: First of all we will take the LCM of the denominator of the left-hand side. And we will solve it. After that we will do the cross multiplication then we will get an equation and will keep the equation to one side and equate it with the zero and we will solve for $x$.

Complete step-by-step answer:
We have
\[\dfrac{1}{{x + 1}} + \dfrac{3}{{5x + 1}} = \dfrac{5}{{x + 4}}\]
Now taking the LCM and solving for it, we get
$ \Rightarrow \dfrac{{5x + 1 + 3x + 1}}{{\left( {x + 1} \right)\left( {5x + 1} \right)}} = \dfrac{5}{{x + 4}}$
On solving the numerator and denominator, we get
$ \Rightarrow \dfrac{{8x + 4}}{{5{x^2} + 6x + 1}} = \dfrac{5}{{x + 4}}$
Now on expanding the equation and doing the cross-multiplication, we get
$ \Rightarrow \left( {8x + 4} \right)\left( {x + 4} \right) = 5\left( {5{x^2} + 6x + 1} \right)$
On solving the multiplication part in both the side, we get
$ \Rightarrow 8{x^2} + 36x + 16 = 25{x^2} + 30x + 5$
On equating the equation and taking all terms on one side, we get
$ \Rightarrow 17{x^2} - 6x - 11 = 0$
Now we will solve the above equation by using the method called factor theorem.
So by splitting the middle term, we will solve the equation, we get
$ \Rightarrow 17{x^2} - 17x + 11x - 11 = 0$
Now taking the common in LHS, we get
$ \Rightarrow 17x\left( {x - 1} \right) + 11\left( {x - 1} \right) = 0$
And it can be written as
$ \Rightarrow \left( {17x + 11} \right)\left( {x - 1} \right) = 0$
Now on equating both the term with zero, we get
$ \Rightarrow x = \left( {\dfrac{{ - 11}}{{17}}} \right)$ Or $x = 1$.
Therefore, $x = \left( {\dfrac{{ - 11}}{{17}}} \right)$ or $x = 1$ will be the value for the $x$.

Additional information:
 Factor theorem is utilized when considering the polynomials. It is a hypothesis that connects variables and zeros of the polynomial. The factor theorem is regularly utilized for figuring a polynomial and finding the foundations of the polynomial. It is an uncommon instance of a polynomial leftover portion hypothesis.

Note: This question can also be solved in different ways. Like we can also use the quadratic formula to solve it. $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$. By using this we can solve any quadratic equations easily. One another method is by using the graphical method.