Solve the linear equation $\dfrac{q+2}{5}=\dfrac{2q-11}{7}$.

Hint: We solve the given linear equation by simplifying the equation. We cross-multiply the equations. The process of cross-multiplication is equivalent to taking the L.C.M of the denominators and multiplying with that. Then we apply the binary operations to get the value of q.

Complete step-by-step solution:
The given equation $\dfrac{q+2}{5}=\dfrac{2q-11}{7}$ is a linear equation of q.
We apply cross-multiplication to multiply $\left( q+2 \right)$ with 7 and $\left( 2q-11 \right)$ with 5.
\[\begin{align}
  & \dfrac{q+2}{5}=\dfrac{2q-11}{7} \\
 & \Rightarrow 7\left( q+2 \right)=5\left( 2q-11 \right) \\
\end{align}\]
We complete the multiplication to get \[7q+14=10q-55\].
Now we apply binary operations to find the solution for q. We take all the variables in one side and the constant terms on the other side.
\[\begin{align}
  & 7q+14=10q-55 \\
 & \Rightarrow 10q-7q=14+55 \\
\end{align}\]
Now we find their solutions
\[\begin{align}
  & 10q-7q=14+55 \\
 & \Rightarrow 3q=69 \\
\end{align}\]
Now we divide both sides with 3 and get
\[\begin{align}
  & 3q=69 \\
 & \Rightarrow q=\dfrac{69}{3}=23 \\
\end{align}\]
We get a value of q as 23 which is the final answer.
Therefore, the solution of $\dfrac{q+2}{5}=\dfrac{2q-11}{7}$ is $q=23$.

Note: The process of cross-multiplication comes from the L.C.M of the denominators. The L.C.M gets multiplied with the both sides of the equation which in turn gives the cross multiplication. In the case of denominators being co-primes, both processes are the same.
For example, if we take $\dfrac{x}{36}=\dfrac{y}{27}$, we can cross multiply to get $27x=36y$ and then we divide both side with their G.C.D value 9.
That gives us
$\begin{align}
  & \dfrac{27x}{9}=\dfrac{36y}{9} \\
 & \Rightarrow 3x=4y \\
\end{align}$
We can also use the L.C.M of the denominators which is 108 and multiply both sides of $\dfrac{x}{36}=\dfrac{y}{27}$ to get the final answer directly.
\[\begin{align}
  & \dfrac{x}{36}\times 108=\dfrac{y}{27}\times 108 \\
 & \Rightarrow 3x=4y \\
\end{align}\].