Question

How do you solve the rational equation $\dfrac{4}{x+2}-\dfrac{3}{x-5}=\dfrac{15}{{{x}^{2}}-3x-10}$?

Answer 1

Hint: In this problem we have to solve the given rational equation that means we need to calculate the value of $x$ for which the given equation is satisfied. We can observe that the equation has arithmetic expressions on both sides of the given equation. So, we will first consider the LHS part. In the LHS we have expressions in denominators, so we will do subtraction of the fractions and simplify the LHS part. After getting the simplified form of the LHS we will equate it to RHS and compare both the values and simplify the equation to get the required answer.

Complete step-by-step solution:
Given expression $\dfrac{4}{x+2}-\dfrac{3}{x-5}=\dfrac{15}{{{x}^{2}}-3x-10}$.
Considering the LHS part of the above equation,
$\Rightarrow LHS=\dfrac{4}{x+2}-\dfrac{3}{x-5}$
Simplifying the above equation by doing subtraction of fractions, then we will get
$\Rightarrow LHS=\dfrac{4\left( x-5 \right)-3\left( x+2 \right)}{\left( x+2 \right)\left( x-5 \right)}$
Applying the distribution law of the multiplication in the above equation, then we will have
$\begin{align}
  & \Rightarrow LHS=\dfrac{4x-4\times 5-3x-3\times 2}{x\left( x-5 \right)+2\left( x-5 \right)} \\
 & \Rightarrow LHS=\dfrac{4x-3x-20-6}{{{x}^{2}}-5x+2x-10} \\
 & \Rightarrow LHS=\dfrac{x-26}{{{x}^{2}}-3x-10} \\
\end{align}$
Equating the above LHS part to the RHS part of the given equation, then we will get
$\Rightarrow \dfrac{x-26}{{{x}^{2}}-3x-10}=\dfrac{15}{{{x}^{2}}-3x-10}$
Comparing the both the values in the above equation, then we will have
$\Rightarrow x-26=15$
Adding $26$ on both sides of the above equation, then we will get
$\begin{align}
  & \Rightarrow x-26+26=15+26 \\
 & \Rightarrow x=41 \\
\end{align}$
Hence the solution of the given equation $\dfrac{4}{x+2}-\dfrac{3}{x-5}=\dfrac{15}{{{x}^{2}}-3x-10}$ is $x=41$.

Note: In this problem we have considered the LHS part and simplified it. We can also consider the RHS part and simplify the denominator which is the quadratic equation. We will use the factorization method and write the quadratic equation in factored form. Now we will compare both the equations after taking LCM on the LHS side. Now we can simplify the obtained equation to get the required result.