How do you solve the system \[-7x-6y=4,x=3y+8\]?

Hint: Assume the given equations as equation (1) and (2) respectively. Now, consider equation (2) where the value of x is given in terms of y. Substitute this value of x in equation (1) and solve for the value of y. Once the value of y is found, substitute it in equation (2) to find the value of x.

Complete step-by-step solution:
Here, we have been provided with the two equations: - \[-7x-6y=4\] and \[x=3y+8\] and we are asked to solve this system, that means we have to find the values of the variables x and y.
Now, let us assume the two given equations as equation (1) and (2) respectively, so we have,
\[\Rightarrow -7x-6y=4\] - (1)
\[\Rightarrow x=3y+8\] - (2)
Let us apply the substitution method to solve the above question. Here, we will substitute the value of the variable x, which is given in equation (2), in equation and solve for the value of y.
Once the value of y is found we will substitute this obtained value in equation (2) to get the value of x.
So, now substituting the value of x given in terms of y from equation (2) in equation (1), we get,
\[\begin{align}
  & \Rightarrow -7\left( 3y+8 \right)-6y=4 \\
 & \Rightarrow -21y-56-6y=4 \\
 & \Rightarrow -27y=56+4 \\
 & \Rightarrow -27y=60 \\
\end{align}\]
Dividing both the sides with -27 and simplifying, we get,
\[\begin{align}
  & \Rightarrow y=\dfrac{-60}{27} \\
 & \Rightarrow y=\dfrac{-20}{9} \\
\end{align}\]
Now, substituting the above obtained values of y in equation (2), we get,
\[\begin{align}
  & \Rightarrow x=3\times \left( \dfrac{-20}{9} \right)+8 \\
 & \Rightarrow x=\dfrac{-20}{3}+8 \\
\end{align}\]
\[\Rightarrow x=\dfrac{-20+24}{3}\]
\[\Rightarrow x=\dfrac{4}{3}\]
Hence, the solution of the given system of equations is given as: - \[\left( x,y \right)=\left( \dfrac{4}{3},\dfrac{-20}{9} \right)\].

Note: One may note that we can also substitute the value of y from equation (1) in equation (2) and then find the value of x first. This will also give the same answer. You may check the answer by substituting the values of x and y obtained, in the given equations. If L.H.S. and R.H.S. turns out to be the same for both the cases then our answer is correct. Remember that we can also solve the system by the elimination method or by the cross – multiplication method. For the cross – multiplication method you must remember the formula otherwise any calculation mistakes may occur.