Question

How do you solve the system of equations $x - y = 0$ and $x - y - 2 = 0$ using substitution?

Answer 1

Hint:A system of two simultaneous equations in two variables is given to us. We have to solve it using a substitution method. For this method, find the value of one variable in the form of another variable from one equation and then put this value in another equation to solve it and determine the variables.

Complete step by step answer:According to the question, a system of two simultaneous equations in two variables is given asking us to use a substitution method to solve it.
The two given equations are:
$
   \Rightarrow x - y = 0{\text{ }}.....{\text{(1)}} \\
   \Rightarrow x - y - 2 = 0{\text{ }}.....{\text{(2)}} \\
 $
For the substitution method, we’ll find the value of one variable in the form of other variable from one equation and put this value in the other equation to find the variables.
So from equation (1), we have:
$
   \Rightarrow x - y = 0 \\
   \Rightarrow x = y \\
 $
Putting $x = y$ in equation (2), we’ll get:
$
   \Rightarrow y - y - 2 = 0 \\
   \Rightarrow - 2 = 0 \\
 $
This condition can never be true as -2 can never be equal to zero. Thus we can safely say that the given system of equations has no solution.
And hence the system of equations is inconsistent.

Note:
If we are given a system of linear equations as shown below:
$
   \Rightarrow {a_1}x + {b_1}y + {c_1} = 0 \\
   \Rightarrow {a_2}x + {b_2}y + {c_2} = 0 \\
 $
There arise three conditions for its solution:
(1) If $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$ then the system of equations will have infinitely many solution and the system is said to be consistent.
(2) If $\dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}}$ then the system of equations will have a unique solution and the system is said to be consistent.
(3) And if $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}$ then the system of equations will have no solution and the system is said to be inconsistent.