Question

How do you solve $\dfrac{3{{x}^{2}}-10x-8}{6{{x}^{2}}+13x+6}\cdot \dfrac{4{{x}^{2}}-4x-15}{2{{x}^{2}}-9x+10}=\dfrac{x-4}{x-2}$?

Answer 1

Hint: We first need to determine the domain set for the given equation. For this, we need to factorise all the quadratic expressions present in the LHS of the given equation using the middle term splitting method. Then setting the denominators equal to zero, we will get all the values which do not belong to the domain set, and hence the domain set of the given equation will be determined. Then by cancelling all the terms common to the numerators and the denominators we will get a simplified equation, which can be easily solved in accordance with the domain set.

Complete step by step solution:
The equation given in the question is
$\dfrac{3{{x}^{2}}-10x-8}{6{{x}^{2}}+13x+6}\cdot \dfrac{4{{x}^{2}}-4x-15}{2{{x}^{2}}-9x+10}=\dfrac{x-4}{x-2}.......(i)$
The LHS of the above equation looks quite cumbersome. So we consider simplifying it.
$\Rightarrow LHS=\dfrac{3{{x}^{2}}-10x-8}{6{{x}^{2}}+13x+6}\cdot \dfrac{4{{x}^{2}}-4x-15}{2{{x}^{2}}-9x+10}$
We can see that the above expression involves quadratic expressions in the numerators and the denominators. For simplifying, we need to factorize them using the middle term splitting method. So we split the middle terms of all of the quadratic expressions to get
$\Rightarrow LHS=\dfrac{3{{x}^{2}}+2x-12x-8}{6{{x}^{2}}+9x+4x+6}\cdot \dfrac{4{{x}^{2}}+6x-10x-15}{2{{x}^{2}}-4x-5x+10}$
Taking terms common from all the quadratic expression, we have
\[\begin{align}
  & \Rightarrow LHS=\dfrac{x\left( 3x+2 \right)-4\left( 3x+2 \right)}{3x\left( 2x+3 \right)+2\left( 2x+3 \right)}\cdot \dfrac{2x\left( 2x+3 \right)-5\left( 2x+3 \right)}{2x\left( x-2 \right)-5\left( x-2 \right)} \\
 & \Rightarrow LHS=\dfrac{\left( 3x+2 \right)\left( x-4 \right)}{\left( 2x+3 \right)\left( 3x+2 \right)}\cdot \dfrac{\left( 2x+3 \right)\left( 2x-5 \right)}{\left( x-2 \right)\left( 2x-5 \right)} \\
\end{align}\]
Now the LHS looks simplified. Let us put it back in the equation (i) to get
\[\Rightarrow \dfrac{\left( 3x+2 \right)\left( x-4 \right)}{\left( 2x+3 \right)\left( 3x+2 \right)}\cdot \dfrac{\left( 2x+3 \right)\left( 2x-5 \right)}{\left( x-2 \right)\left( 2x-5 \right)}=\dfrac{x-4}{x-2}.......(ii)\]
We know that the denominator of a number can never be equal to zero. Therefore from the above equation we can write
$\begin{align}
  & \Rightarrow 2x+3\ne 0 \\
 & \Rightarrow x\ne -\dfrac{3}{2} \\
\end{align}$
\[\begin{align}
  & \Rightarrow 3x+2\ne 0 \\
 & \Rightarrow x\ne -\dfrac{2}{3} \\
\end{align}\]
$\begin{align}
  & \Rightarrow x-2\ne 0 \\
 & \Rightarrow x\ne 2 \\
\end{align}$
And
$\begin{align}
  & \Rightarrow 2x-5\ne 0 \\
 & \Rightarrow x\ne \dfrac{5}{2} \\
\end{align}$
Therefore, the domain set for the given equation becomes $x\in R-\left\{ -\dfrac{3}{2},-\dfrac{2}{3},2,\dfrac{5}{2} \right\}$
Now, we again consider the equation (ii) to get
\[\begin{align}
  & \Rightarrow \dfrac{\left( 3x+2 \right)\left( x-4 \right)}{\left( 2x+3 \right)\left( 3x+2 \right)}\cdot \dfrac{\left( 2x+3 \right)\left( 2x-5 \right)}{\left( x-2 \right)\left( 2x-5 \right)}=\dfrac{x-4}{x-2} \\
 & \Rightarrow \dfrac{\left( x-4 \right)}{\left( 2x+3 \right)}\cdot \dfrac{\left( 2x+3 \right)}{\left( x-2 \right)}=\dfrac{x-4}{x-2} \\
 & \Rightarrow \dfrac{\left( x-4 \right)}{\left( x-2 \right)}=\dfrac{x-4}{x-2} \\
\end{align}\]
So we have obtained LHS equal to RHS. This means that the given equation is satisfied for the complete domain set, which we obtained above as $x\in R-\left\{ -\dfrac{3}{2},-\dfrac{2}{3},2,\dfrac{5}{2} \right\}$.
Hence, the solution of the given equation is $x\in R-\left\{ -\dfrac{3}{2},-\dfrac{2}{3},2,\dfrac{5}{2} \right\}$.

Note:
If we simplify the LHS of the given equation directly without determining the domain set, then we will obtain the LHS equal to RHS, as obtained in the above solution. So we may conclude that the given equation is not just an equation, but an identity. But it is important to note that an identity is an equation which is satisfied for each real value of the variable, that is, for $x\in R$. But from the domain set, we can see that the values $-\dfrac{3}{2},-\dfrac{2}{3},2,\dfrac{5}{2}$ are not allowed. So the given equation is not an identity and thus we must not forget to determine the domain set before simplification.