Question

How do you solve the rational equation $\dfrac{{{\text{x + 1}}}}{{{\text{x - 1}}}} = \dfrac{{\text{x}}}{3} + \dfrac{2}{{{\text{x - 1}}}}$ ?

Answer 1

Hint: Here in this question they have given a rational equation and asked us to solve it. First we will add the right hand side by cross multiplying and derive an expression and then we have to equate RHS and LHS by cross multiplying and form an equation, with that we will be able to solve the equation by finding the value of ${\text{x}}$.

Complete step-by-step solution:
The given rational equation is $\dfrac{{{\text{x + 1}}}}{{{\text{x - 1}}}} = \dfrac{{\text{x}}}{3} + \dfrac{2}{{{\text{x - 1}}}}$, we need to solve this.
First, we will solve the RHS, $\dfrac{{{\text{x + 1}}}}{{{\text{x - 1}}}} = \dfrac{{\text{x}}}{3} + \dfrac{2}{{{\text{x - 1}}}}$
By cross multiplying the RHS, we can find the sum of the given terms.
For that we have to multiply both the denominator and then cross multiply the numerator with the another denominator, like this,
$ \Rightarrow \dfrac{{{\text{x + 1}}}}{{{\text{x - 1}}}} = \dfrac{{{\text{x}}\left( {{\text{x - 1}}} \right) + 2(3)}}{{3\left( {{\text{x - 1}}} \right)}}$
Let us simplify we get,
$ \Rightarrow \dfrac{{{\text{x + 1}}}}{{{\text{x - 1}}}} = \dfrac{{{\text{x}}\left( {{\text{x - 1}}} \right) + 6}}{{3\left( {{\text{x - 1}}} \right)}}$
We are keeping $\left( {{\text{x - 1}}} \right)$ unbothered, because there is another $\left( {{\text{x - 1}}} \right)$ in the LHS too. It can help us cancel the terms. Now, cross multiplying the RHS and LHS we get,
\[ \Rightarrow \left( {{\text{x + 1}}} \right)\left( {{\text{3x - 3}}} \right) = \left( {{{\text{x}}^2}{\text{ - x}} + 6} \right)\left( {{\text{x - 1}}} \right)\]
As we can see there is $\left( {{\text{x - 1}}} \right)$ on both sides, so we can cancel it.
\[ \Rightarrow \left( {3{x^2} + 3x - 3x - 3} \right) = \left( {{{\text{x}}^3}{\text{ - }}{{\text{x}}^2}{\text{ + 6x - }}{{\text{x}}^2}{\text{ + x - }}6} \right)\]
On simplification we get,
\[ \Rightarrow {\text{3}}{{\text{x}}^2}{\text{ - 3}} = {{\text{x}}^3}{\text{ - 2}}{{\text{x}}^2} + 7x - 6\]
Now we have to separate all the terms to one side,
\[ \Rightarrow {{\text{x}}^3}{\text{ - 2}}{{\text{x}}^2}{\text{ + 7x - }}6{\text{ - 3}}{{\text{x}}^2}{\text{ + 3}} = 0\]
\[ \Rightarrow {{\text{x}}^3}{\text{ - 5}}{{\text{x}}^2}{\text{ + 7x - 3}} = 0\]
\[\begin{array}{*{20}{c}}
  {1\left| \!{\underline {\,
  \begin{gathered}
  1\,\,\,\, - 5\,\,\,\,\,7\,\,\,\, - 3 \\
  0\,\,\,\,\,\,\,\,1\,\,\, - 4\,\,\,\,0 \\
\end{gathered} \,}} \right. } \\
  {\,\,\,\,\,\,\,1\,\,\,\,\, - 4\,\,\,\,\,3\,\,\,\,\,\,\left| \!{\underline {\,
  0 \,}} \right. \,\,\,\,}
\end{array}\]
After solving this, we get a quadratic equation.
\[ \Rightarrow \left( {{\text{x - 1}}} \right){\text{(}}{{\text{x}}^2}{\text{ - 4x + 3)}} = 0...\left( 1 \right)\]
Now, we need to find the roots of the equation i.e. the value of ${\text{x}}$ , we will do that by splitting the middle term method.
In this method we have to split the middle term (i.e. ${\text{bx}}$ ) in such a way that it should be a sum or a difference of one of the factors of ${\text{ac}}$ .
Here, we split ${\text{ - 4x}}$ as (\[{\text{ - 3x - x}}\]) which are the factors of ${\text{ac}}$(i.e. $ - 3, - 1$ )
\[ \Rightarrow {{\text{x}}^2}{\text{ - 3x - x + 3}} = 0\]
Now we have to take the commons out, clearly \[{\text{(x - 2)}}\]is common in both the terms.
\[ \Rightarrow {\text{x(x - 3) - 1(x - 3)}} = 0\]
We can take \[{\text{(x - 3)}}\] out and we get,
\[ \Rightarrow {\text{(x - 3) (x - 1)}} = 0\]
Solving this we get,
\[ \Rightarrow {\text{(x - 3) = 0, (x - 1)}} = 0\]
$ \Rightarrow {\text{x = 3 , x = 1}}$
Then equation becomes, $x = 3,1,1$

Therefore the value of ${\text{x is 3,1}}$

Note: Alternative method to solve x:
After forming an equation, we can solve that equation in another method too, by using the quadratic equation formula:
${\text{x = }}\dfrac{{{{ - b \pm }}\sqrt {{{\text{b}}^{\text{2}}}{\text{ - 4ac}}} }}{{{\text{2a}}}}$
The equation is \[{{\text{x}}^2}{\text{ - 4x + 3}} = 0\]
Here ${\text{a = 1, b = - 4 , c = 3}}$
Applying in the formula,
$ \Rightarrow {\text{x = }}\dfrac{{ - ( - 4) \pm \sqrt {{{( - 4)}^2} - 4(1)(3} )}}{{2(1)}}$
\[ \Rightarrow {\text{x = }}\dfrac{{4 \pm \sqrt {16 - 12} }}{2}\]
\[ \Rightarrow {\text{x = }}\dfrac{{4 \pm \sqrt 4 }}{2}\]
\[ \Rightarrow {\text{x = }}\dfrac{{4 \pm 2}}{2}\]
Now we can split it into two, as the expression involves $ \pm $ in it, one can be plus the other can be minus.
\[ \Rightarrow {\text{x = }}\dfrac{{4 + 2}}{2},\dfrac{{4 - 2}}{2}\]
On simplify we get
\[ \Rightarrow {\text{x = }}\dfrac{6}{2},\dfrac{2}{2}\]
On divide the term and we get,
$ \Rightarrow {\text{x = 3, x = 1}}$
Therefore the value of ${\text{x is 3,1}}$