Question

How do you solve the system of equations: $ y = 2x + 1 $ and $ y = 4x - 1 $ ?

Answer 1

Hint: The given system contains two linear equations in two variables. So, they can be solved, to get either a unique solution, infinitely many solutions or no solution. There are various methods available for solving such equations like the substitution method, Crammer’s Rule etc.

Complete step by step solution:
We know that two linear equations $ {a_1}x + {b_1}y = {c_1}{\text{ and }}{a_2}x + {b_2}y = {c_2} $ in two variables will have,
(i) exactly one solution if $ \dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}} $ .
(ii) infinitely many solutions if $ \dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}} $ .
(ii) and no solution if $ \dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}} $ .
The system of linear equations given are,
 $ y = 2x + 1 $ - - - - - - - - - - - - - - - - - - - - - (1)
 $ y = 4x - 1 $ - - - - - - - - - - - - - - - - - - - - - (2)
Now, on comparing (1) and (2) with the standard equations we get $ {a_1} = - 2,{b_1} = 1,{c_1} = 1 $ and $ {a_2} = - 4,{b_2} = 1,{c_2} = - 1 $ .
Here, we observe that $ \dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}} $ as $ \dfrac{1}{2} \ne 1 $ . So, the given system is consistent and has exactly one solution.
Now, to solve the given system we will use the method of substitution.
Since, the value of $ y $ is readily available in equation (1), we will substitute the $ y $ in equation (2) with the RHS of (1).
 $ \Rightarrow 2x + 1 = 4x - 1 $
Now, on bringing the $ x $ terms to one side, we get
 $ \Rightarrow 2x - 4x = - 1 - 1 = - 2 $
On further simplification we get,
 $ \Rightarrow - 2x = - 2 $
We will divide both sides by $ - 2 $ .
 $ \Rightarrow x = 1 $
Now, to find the value of $ y $ we can substitute $ x = 1 $ in any of the two equations.
We will substitute $ x = 1 $ in equation (1) to get, $ y = 2(1) + 1 $
 $ \Rightarrow y = 2 + 1 = 3 $
Hence, the solution of the given equations $ y = 2x + 1 $ and $ y = 4x - 1 $ is $ x = 1{\text{ and }}y = 3 $ .
So, the correct answer is “$ x = 1{\text{ and }}y = 3 $”.

Note: Geometric interpretation would be that the lines representing these equations intersect at the point $ \left( {1,3} \right) $ . Also, the two lines representing the system would be parallel if they do not have any solution and will coincide if they have infinitely many solutions.