Solve the following equations by the substitution method:
(1) \[x+y=14\] and \[x-y=4\]
(2) \[s-t=3\] and \[\dfrac{s}{3}+\dfrac{t}{2}=6\]

Hint: We solve this problem by using the substitution method. The substitution method is the method used to solve linear equations of two variables by substituting one of the equations in another equation. This means that we find the value of one of the variables in terms of another variable using one equation and substitute in another equation to find the solution.

Complete step-by-step solution
We are given two sets of linear equations of two variables
Let us find the solution one by one.
(1) \[x+y=14\] and \[x-y=4\]
We know that the substitution method is the method used to solve linear equations of two variables by substituting one of the equations in another equation.
We know that in a substitution method we find the value of one of the variables in terms of another variable using one equation and substitute in another equation to find the solution.
Let us take the first equation and find the value of \['y'\] in terms of \['x'\] then we get
\[\begin{align}
  & \Rightarrow x+y=14 \\
 & \Rightarrow y=14-x \\
\end{align}\]
Now, by substituting the value of \['y'\] in second equation we get
\[\begin{align}
  & \Rightarrow x-\left( 14-x \right)=4 \\
 & \Rightarrow 2x=14+4 \\
 & \Rightarrow x=9 \\
\end{align}\]
Now, by substituting the value \[x=9\] in the value of \['y'\] we get
\[\Rightarrow y=14-9=5\]
Therefore we can conclude that the solution of this set as \[x=9,y=5\]
(2) \[s-t=3\] and \[\dfrac{s}{3}+\dfrac{t}{2}=6\]
Now, let us take the first equation and find the value of \['s'\] in terms of \['t'\] then we get
\[\begin{align}
  & \Rightarrow s-t=3 \\
 & \Rightarrow s=t+3 \\
\end{align}\]
Now, by substituting the value of \['s'\] in second equation we get
\[\begin{align}
  & \Rightarrow \dfrac{t+3}{3}+\dfrac{t}{2}=6 \\
 & \Rightarrow \dfrac{t}{3}+\dfrac{t}{2}=6-1=5 \\
\end{align}\]
Now, by adding the terms in the above equation using the LCM we get
\[\begin{align}
  & \Rightarrow \dfrac{2t+3t}{6}=5 \\
 & \Rightarrow \dfrac{5t}{6}=5 \\
 & \Rightarrow t=6 \\
\end{align}\]
Now, by substituting the value \[t=6\] in the value of \['s'\] we get
\[\Rightarrow s=6+3=9\]
Therefore we can conclude that the solution of this set of equations is \[s=9,t=6\]

Note: We can solve the linear equations of two variables by using the elimination method. That is we add or subtract the given two equations after multiplying with a certain number so that we can get one term same in two equations.
Let us solve the first set using the elimination method.
(1) \[x+y=14\] and \[x-y=4\]
Here we can see that the coefficient of \['x'\] in both equations is same
So, by subtracting second equation from first equation we get
\[\begin{align}
  & \Rightarrow \left( x+y \right)-\left( x-y \right)=14-4 \\
 & \Rightarrow 2y=10 \\
 & \Rightarrow y=5 \\
\end{align}\]
Now, by substituting the value of \['y'\] in first equation we get
\[\begin{align}
  & \Rightarrow x+5=14 \\
 & \Rightarrow x=9 \\
\end{align}\]
Therefore we can conclude that the solution of this set as \[x=9,y=5\]
Here we can see that the solution in this method is the same as the solution in the substitution method.
But in the question, we are asked to find the solution using the substitution method so using the elimination method is not correct for this problem.
But we can use any method if we are not mentioned to use a specific method.