How do you solve $ 6x + y = 13 $ and $ y - x = - 8 $ using substitution?

Hint: In order to determine the solution of a given system of equations having two variables using the substitution method, find the value of one variable $ y $ in terms of another variable $ x $ from the 2nd equation and substitute $ y $ in the 1st equation. Solve it to obtain the value of $ x $ and put this value in any of the given equations to obtain the value of $ y $ .

Complete step by step solution:
 We are given pair of linear equations in two variables $ 6x + y = 13 $ and $ y - x = - 8 $
 $ 6x + y = 13 $ ---(1)
 $ y - x = - 8 $ ----(2)
In order to solve the system of equations, we have many methods like, substitution, elimination of term, and cross-multiplication.
Here we will be using a substitution method, by finding the value of one variable in terms of another variable from any equation and substitute that variable in the another equation and solving it .
We will be deriving the expression for variable $ y $ from the equation (2)
 $ y - x = - 8 $
 $ y = - 8 + x $ ----(3)
Now substituting the variable $ y $ in the equation (1), by using the above result. We get equation (2) as
 $
  6x + \left( { - 8 + x} \right) = 13 \\
  6x - 8 + x = 13 \;
  $
Now combining all the like terms and rearranging the above equation such that term having $ x $ are on the LHS and constant terms are on the RHS by transposing the terms.
 $
\Rightarrow 6x + x = 13 + 8 \\
\Rightarrow 7x = 21 \;
  $
Dividing both sides of the equation with the coefficient of $ x $ i.e. $ 7 $
 $ \dfrac{{7x}}{7} = \dfrac{{21}}{7} $
 $\Rightarrow x = 3 $ ---(3)
Now, putting the obtained value of $ x $ in the equation (3), we get
 $
\Rightarrow y = - 8 + 3 \\
\Rightarrow y = - 5 \;
  $
Therefore, the solution of system of given equations is $ x = 3,y = - 5 $
So, the correct answer is “ $ x = 3,y = - 5 $ ”.

Note: 1. One must be careful while calculating the answer as calculation error may come.
2. Solution of two linear equations can be done by using elimination method, substitution method and cross multiplication method.
3.It’s not compulsory to always substitute the term with $ y $ . We have to judge according to the equation whose variable substitution cost is less.