Question

How do you multiply $ \dfrac{x}{{{x^2} - 1}} + \dfrac{2}{{x + 1}} = 1 + \dfrac{1}{{2x - 2}} $ ?

Answer 1

Hint: In order to multiply the above given rational equation having variable $ x $ ,first rewrite the equation using the identity $ {A^2} - {B^2} = \left( {A - B} \right)\left( {A + B} \right) $ , then multiply both sides of the equation with $ \left( 2 \right)\left( {x - 1} \right)\left( {x + 1} \right) $ . Expand the terms using distributive property of multiplication $ A\left( {B + C} \right) = AB + AC $ . Combine all the like terms by transposing all the term on one side to obtain the required result.

Complete step by step solution:
We are given a rational equation in variable $ x $
 $ \dfrac{x}{{{x^2} - 1}} + \dfrac{2}{{x + 1}} = 1 + \dfrac{1}{{2x - 2}} $
Rewriting the above equation by using the identity $ {A^2} - {B^2} = \left( {A - B} \right)\left( {A + B} \right) $ in the first term in the LHS of the equation and taking common from the denominator in the RHS , our equation becomes
 $ \dfrac{x}{{\left( {x - 1} \right)\left( {x + 1} \right)}} + \dfrac{2}{{x + 1}} = 1 + \dfrac{1}{{2\left( {x - 1} \right)}} $
Simplifying the above equation by multiplying both sides of the equation with $ \left( {x - 1} \right)\left( {x + 1} \right) $ , we get
 $ \left( {x - 1} \right)\left( {x + 1} \right)\left( {\dfrac{x}{{\left( {x - 1} \right)\left( {x + 1} \right)}} + \dfrac{2}{{x + 1}}} \right) = \left( {x - 1} \right)\left( {x + 1} \right)\left( {1 + \dfrac{1}{{2\left( {x - 1} \right)}}} \right) $
For the multiplication and expansion of the terms , use the distributive law of multiplication as $ A\left( {B + C} \right) = AB + AC $
Simplifying it further , we have
\[
\Rightarrow x + \left( {x - 1} \right) \times 2 = \left( {x - 1} \right)\left( {x + 1} \right)1 + \left( {x + 1} \right) \times \dfrac{1}{2} \\
\Rightarrow x + 2x - 2 = {x^2} - 1 + \dfrac{x}{2} + \dfrac{1}{2} \;
 \]
Multiplying both sides of the equation with number $ 2 $ in order to not have any rational term
\[
\Rightarrow 2\left( {x + 2x - 2} \right) = 2\left( {{x^2} - 1 + \dfrac{x}{2} + \dfrac{1}{2}} \right) \\
\Rightarrow 2x + 4x - 4 = 2{x^2} - 2 + x + 1 \;
 \]
Now combining all the terms, and rearranging the above quadratic equation in the form of standard quadratic equation as $ a{x^2} + bx + c $
For combining all the like term, first transpose all the terms from left-hand side to right-hand side
\[0 = 2{x^2} - 2 + x + 1 - 2x - 4x + 4\]
Combining terms, we get
\[2{x^2} - 5x + 3 = 0\]
Therefore the multiplied form of the given equation is \[2{x^2} - 5x + 3 = 0\].
So, the correct answer is “ \[2{x^2} - 5x + 3 = 0\]”.

Note: Quadratic Equation: A quadratic equation is a equation which can be represented in the form of $ a{x^2} + bx + c $ where $ x $ is the unknown variable and a,b,c are the numbers known where $ a \ne 0 $ .If $ a = 0 $ then the equation will become linear equation and will no more quadratic .
The degree of the quadratic equation is of the order 2.