How do you solve $ x = y + 3 $ and $ x - {y^2} = 3y $ using substitution?

Hint: To solve the given equations, first we will substitute the first equation in the second equation to solve for the one variable. Again substitute the found variable in the first equation to solve for the other variable.

Complete step by step solution:
Given equations-
 $ x = y + 3 $ ….eq.(i)
 $ x - {y^2} = 3y $ …eq.(ii)
Now, we will substitute eq.(i) in eq.(ii), then we get-
 $
x - {y^2} = 3y \\
   \Rightarrow (y + 3) - {y^2} = 3y \\
   \Rightarrow {y^2} + 2y - 3 = 0 \\
   \Rightarrow (y - 1)(y + 3) = 0 \;
 $
Now, we will find the value of $ y $ :
 $
  (y - 1) = 0\,\,,\,(y + 3) = 0 \\
   \Rightarrow y = 1\,\,,\,\,y = - 3 \;
 $
Now, in eq.(i) , we will put the above values to find the values of $ x $ :
If $ y = 1 $ , then $ x = y + 3 = 1 + 3 = 4 $
If $ y = - 3 $ , then $ x = y + 3 = - 3 + 3 = 0 $
Hence, the solutions are $ x = 4,y = 1 $ and $ x = 0,y = - 3 $ .
In coordinate form:
 $ (4,1)\,and\,(0, - 3) $ .
So, the correct answer is “ $ (4,1)\,and\,(0, - 3) $ ”.

Note: Steps for applying the substitution method:-
Step-1: Solve one equation for one of the variables.
Step-2: Substitute (plug-in) this expression into the other equation and solve.
Step-3: Substitute the value into the original equation to find the corresponding variable.