Question

How do you solve \[\dfrac{1}{{x - 1}} + \dfrac{3}{{x + 1}} = 2\] ?

Answer 1

Hint: A rational equation is an equation containing at least one rational expression. Rational expressions typically contain a variable in the denominator. Solve rational equations by clearing the fractions by multiplying both sides of the equation by the least common denominator (LCD). Multiply
the terms inside the brackets. Simplify the term by addition and multiplication operations.

Complete step-by-step solution:
The given rational equation is
\[ \Rightarrow \dfrac{1}{{x - 1}} + \dfrac{3}{{x + 1}} = 2\]
To solve the equation, we have to clear denominators now, for that we have to do cross multiplication in the equation
$ \Rightarrow \dfrac{{1 \times \left( {x + 1} \right) + \,3 \times \left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = 2$
Now we can reduce the equation by multiplying the numbers inside the brackets.
Therefore, we get
$\dfrac{{x + 1 + 3x - 3}}{{{x^2} + x - x - 1}} = 2$
Now we can reduce the $x$ term by adding $3x$ and $x$ in numerator and cancelling $ + x$ and $ - x$ in denominator
$ \Rightarrow \dfrac{{4x - 2}}{{{x^2} - 1}} = 2$
Bring the denominator to the right hand side, we get
$ \Rightarrow \left( {4x - 2} \right) = 2\left( {{x^2} - 1} \right)$
Multiply inside brackets on RHS, we get
$ \Rightarrow \left( {4x - 2} \right) = \left( {2{x^2} - 2} \right)$
Now divide both sides by the value $2$, we get
$ \Rightarrow \left( {\dfrac{{4x - 2}}{2}} \right) = \left( {\dfrac{{2{x^2} - 2}}{2}} \right)$
On solving the above equation, we get
$ \Rightarrow \left( {2x - 1} \right) = \left( {{x^2} - 1} \right)$
Bringing same terms one side we get,
$ \Rightarrow {x^2} - 2x = - 1 + 1$
Now reducing the equation further, we will get
$ \Rightarrow {x^2} - 2x = 0$
From the above equation taking $x$ common, we get
$x\left( {x - 2} \right) = 0$
Separately equate each term to zero,
$x = 0$ and
$ \Rightarrow x - 2 = 0$
From this we will get the solution of the equation
That is
${x_1} = 0$ and ${x_2} = 2$

Therefore $x$ can take the values 0 and 2.

Note: Clearing denominators in rational equations is also known as clearing fraction in an equation. There are
Many times when a problem becomes easier to solve if you don’t have to worry about adding and
Subtracting fractions. To clear the denominators you will need to multiply both sides of the equation by
the smallest number both denominators divide evenly into. In order to clear denominators we need to
take LCM. Finding the LCD of a list of values is the same as finding the LCM of
the denominators of those values $x - 1$ and $x + 1$. Since $x - 1$ and $x + 1$ contain variables, there are
Four steps to find the LCM. If there numeric also present in the problems, find LCM for
the numeric, variable, and compound variable parts. Then, multiply them all together.
Steps to find the LCM are:
$1)$ Find the LCM for the numeric part
$2)$ Find the LCM for the variable part
$3)$ Multiply each LCM together.
The LCM is the smallest positive number that all of the numbers divide into evenly.
$1)$ List the prime factors of each number.
$2)$ Multiply each factor the greatest number of times it occurs in either number.