Solve the equations \[4x+\dfrac{6}{y}=15\] and \[6x-\dfrac{8}{y}=14\] also find the value of \['p'\] if \[y=px-2\]

Hint: We solve this problem by using the elimination method. In the elimination method, we try to eliminate one of the variables from the two equations by adding or subtracting the equations after getting the same term in two equations. If we can eliminate one variable then we get an equation of a single variable which will be easy to find the required values.

Complete step-by-step solution
We are given that the first equation as
\[\Rightarrow 4x+\dfrac{6}{y}=15....equation(i)\]
We are given that the second equation as
\[\Rightarrow 6x-\dfrac{8}{y}=14\]
Here, we can see that the coefficients of all terms are even numbers
So, by dividing the terms with 2 into both sides we get
\[\Rightarrow 3x-\dfrac{4}{y}=7.....equation(ii)\]
Here we can see that the coefficients of \['x'\] in two equations are 4 and 3 respectively.
So, let us multiply the equation (i) with 3 and equation (ii) with 4 then by subtracting equation (ii) from equation (i) we get
\[\Rightarrow 3\left( 4x+\dfrac{6}{y} \right)-4\left( 3x-\dfrac{4}{y} \right)=\left( 3\times 15 \right)-\left( 4\times 7 \right)\]
Now, by multiplying the terms in the above equation and adding them together we get
\[\begin{align}
  & \Rightarrow 12x+\dfrac{18}{y}-12x+\dfrac{16}{y}=45-28 \\
 & \Rightarrow \dfrac{34}{y}=17 \\
 & \Rightarrow y=2 \\
\end{align}\]
Now, let us substitute the value of \['y'\] in equation (i) we get
\[\begin{align}
  & \Rightarrow 4x+\dfrac{6}{2}=15 \\
 & \Rightarrow 4x=15-3 \\
 & \Rightarrow x=3 \\
\end{align}\]
Therefore, the solution of given two equations is \[x=3,y=2\]
We are given that the third equation that is
\[y=px-2\]
Now, by substituting \[x=3,y=2\] in above equation we get
\[\begin{align}
  & \Rightarrow 2=3p-2 \\
 & \Rightarrow 3p=4 \\
 & \Rightarrow p=\dfrac{4}{3} \\
\end{align}\]
Therefore, the value of \['p'\] is $\dfrac{4}{3} $.

Note: We can find the solution of given two equations by using the substitution method.
We are given that the first equation as
\[\Rightarrow 4x+\dfrac{6}{y}=15....equation(i)\]
We are given that the second equation as
\[\begin{align}
  & \Rightarrow 6x-\dfrac{8}{y}=14 \\
 & \Rightarrow \dfrac{8}{y}=6x-14 \\
 & \Rightarrow \dfrac{1}{y}=\dfrac{1}{8}\left( 6x-14 \right)....equation(ii) \\
\end{align}\]
Now, by substituting equation (ii) in equation (i) we get
\[\begin{align}
  & \Rightarrow 4x+6\left( \dfrac{1}{8}\left( 6x-14 \right) \right)=15 \\
 & \Rightarrow 4x+\dfrac{9x}{2}-\dfrac{21}{2}=15 \\
 & \Rightarrow \dfrac{17x}{2}=\dfrac{51}{2} \\
 & \Rightarrow x=3 \\
\end{align}\]
Now by substituting the value of \['x'\] in equation (i) we get
\[\begin{align}
  & \Rightarrow 4\times 3+\dfrac{6}{y}=15 \\
 & \Rightarrow \dfrac{6}{y}=3 \\
 & \Rightarrow y=2 \\
\end{align}\]
Therefore, the solution of given two equations is \[x=3, y=2\]