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

Hint: For the given question we are given to solve the equation \[\dfrac{2x}{x+3}=\dfrac{3x}{x-3}\]. For that we have to observe the equation that we could solve the equation by examining the equation. To solve the given question, we have moved the all the equation to the right hand side.

Complete step by step solution:
The given equation is \[\dfrac{2x}{x+3}=\dfrac{3x}{x-3}\].
We have to move \[\dfrac{3x}{x-3}\] to left hand side of the equation by subtracting both sides
\[\begin{align}
  & \Rightarrow \dfrac{2x}{x+3}-\dfrac{3x}{x-3}=0 \\
 & \Rightarrow \dfrac{2x\left( x-3 \right)-3x\left( x+3 \right)}{\left( x+3 \right)\left( x-3 \right)}=0 \\
\end{align}\]
Now doing cross multiplication we get
\[\begin{align}
  & \Rightarrow 2x\left( x-3 \right)-3x\left( x+3 \right)=0 \\
 & \Rightarrow x\left( 2x-6-3x-9 \right)=0 \\
 & \Rightarrow -x\left( x+15 \right)=0 \\
\end{align}\]
Now we have to move negative
\[\Rightarrow x\left( x+15 \right)=0\]
The multiplication of two terms gives 0 which means the solutions are $x=0,-15$.

Therefore, the numerical $x=0,-15$ is the exact solution of the given problem \[\dfrac{2x}{x+3}=\dfrac{3x}{x-3}\].

Note: We can do this problem even by using distributive law. Students should be careful while doing the cross multiplication. If we want to check the answer for verification substitute $x=0$ in the equation and check it.
 We can also directly cross-multiply \[\dfrac{2x}{x+3}=\dfrac{3x}{x-3}\] to get
\[\begin{align}
  & \dfrac{2x}{x+3}=\dfrac{3x}{x-3} \\
 & \Rightarrow 2{{x}^{2}}-6x=3{{x}^{2}}+9x \\
 & \Rightarrow {{x}^{2}}+15x=0 \\
\end{align}\]
We solve the quadratic equation using quadratic solving
We know for a general equation of quadratic $a{{x}^{2}}+bx+c=0$, the value of the roots of x will be $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
In the given equation we have ${{x}^{2}}+15x=0$. The values of a, b, c are $1,15,0$ respectively.
We put the values and get x as $x=\dfrac{-15\pm \sqrt{{{15}^{2}}-4\times 1\times 0}}{2\times 1}=\dfrac{-15\pm \sqrt{225}}{2}=\dfrac{-15\pm 15}{2}=0,-15$.