Question

How do you solve \[\dfrac{3x+6}{{{x}^{2}}-4}=\dfrac{x+1}{x-2}\]?

Answer 1

Hint: To solve \[\dfrac{3x+6}{{{x}^{2}}-4}=\dfrac{x+1}{x-2}\], we first have to consider the LHS. Take 3 common from the numerator and use the identity \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\] to simplify the denominator. Here, a=x and b=2. Cancel the common term (x+2) from the numerator and denominator. Equate it with the RHS. Multiply the whole equation with (x-2) and cancel the common term from the numerator and the denominator. Then, subtract the obtained equation by 1 to find the value of x.

Complete step by step answer:
According to the question, we are asked to solve \[\dfrac{3x+6}{{{x}^{2}}-4}=\dfrac{x+1}{x-2}\].
We have been given the equation is \[\dfrac{3x+6}{{{x}^{2}}-4}=\dfrac{x+1}{x-2}\]. --------(1)
First, we have to consider the LHS.
We find that 3 are common in both the terms of the numerator. On taking 3 out of the bracket, we get
\[\dfrac{3x+6}{{{x}^{2}}-4}=\dfrac{3\left( x+2 \right)}{{{x}^{2}}-4}\].
\[\Rightarrow \dfrac{3x+6}{{{x}^{2}}-4}=\dfrac{3\left( x+2 \right)}{{{x}^{2}}-{{2}^{2}}}\]
We know that \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\]. Let us use this identity in the denominator of the equation. Here, a=x and b=2. On simplification, we get
\[\dfrac{3x+6}{{{x}^{2}}-4}=\dfrac{3\left( x+2 \right)}{\left( x-2 \right)\left( x+2 \right)}\]
We find that (x+2) is common in both the numerator and denominator. On cancelling the common term, we get
\[\dfrac{3x+6}{{{x}^{2}}-4}=\dfrac{3}{\left( x-2 \right)}\]
Now, let us equate the simplified LHS with the RHS of the given equation (1).
\[\Rightarrow \dfrac{3}{x-2}=\dfrac{x+1}{x-2}\]
Let us multiply the equation with (x-2) on both sides of the equation. We get
\[\dfrac{3}{x-2}\times \left( x-2 \right)=\dfrac{x+1}{x-2}\times \left( x-2 \right)\]
We find that (x-2) are common in both the numerator and denominator of the LHS and RHS.
On cancelling the common terms, we get
\[3=x+1\]
\[\Rightarrow x+1=3\]
Now, subtract 1 from both the sides of the equation.
\[\Rightarrow x+1-1=3-1\]
We know that terms with the same magnitude and opposite signs cancel out.
\[\Rightarrow x=3-1\]
\[\therefore x=2\]

Hence, the value of x in the equation \[\dfrac{3x+6}{{{x}^{2}}-4}=\dfrac{x+1}{x-2}\] is 2.

Note: We can also solve this question by cross multiplying the given equation.
Therefore, we get
\[\left( {{x}^{2}}-4 \right)\left( x+1 \right)=\left( 3x+6 \right)\left( x-2 \right)\]
Take the common terms out of the bracket and use the identity \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\].
Then, make necessary calculations and find the value of x as obtained in the above solution.