Question: How do you determine whether the given ordered pair \[(2, - 3)\] is a solution of the system \[x = 2...

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

Solution

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.