Question

How do you determine whether the given ordered pair \[(2, - 3)\] is a solution of the system \[x = 2y + 8\] and \[2x + y = 1\]?

Answer 1

Hint: We use the concept of the ordered pair given to us i.e. the general ordered pair is of the form \[(x,y)\] and we write the value of x and y from comparing given ordered pair to general ordered pair. Calculate each side of the equation by substituting the values of x and y. If the left side of the equation comes out to be equal to the right side then the ordered pair is a solution of the equation.

Complete step-by-step answer:
We are given two equations:
\[x = 2y + 8\] … (1)
\[2x + y = 1\] … (2)
We are given the ordered pair \[(2, - 3)\]
We have to check if the ordered pair \[(2, - 3)\] is a solution of the system of linear equations (1) and (2)
For an ordered pair to be a solution of a system of linear equations, it has to satisfy the two linear equations separately.
If we compare the given ordered pair i.e. \[(2, - 3)\] to general ordered pair \[(x,y)\], we can write
\[x = 2,y = - 3\]
We will substitute the values of ‘x’ and ‘y’ in each equation one by one and check if LHS is equal to RHS of the equation.
For equation (1):
Substitute \[x = 2,y = - 3\] in equation (1)
\[ \Rightarrow 2 = 2 \times ( - 3) + 8\]
Solve RHS of the equation
\[ \Rightarrow 2 = - 6 + 8\]
\[ \Rightarrow 2 = 2\]
\[\because \]LHS \[ = \] RHS
\[\therefore \]Ordered pair \[(2, - 3)\] is a solution of the equation \[x = 2y + 8\]
For equation (2):
Substitute \[x = 2,y = - 3\] in equation (1)
\[ \Rightarrow 2 \times (2) + ( - 3) = 1\]
Solve RHS of the equation
\[ \Rightarrow 4 - 3 = 1\]
\[ \Rightarrow 1 = 1\]
\[\because \]LHS \[ = \] RHS
\[\therefore \]Ordered pair \[(2, - 3)\] is a solution of the equation \[2x + y = 1\]

Since the ordered pair \[(2, - 3)\] is a solution for both the linear equations, then the ordered pair \[(2, - 3)\] will be a solution for the system of equations \[x = 2y + 8\] and \[2x + y = 1\].

Note:
Many students make mistakes when they try to write the system in the form of a matrix. Then they try to solve the matrix which is wrong. Here we don’t need to convert in matrix form as we only have 2 equations in 2 variables.