### Theory:

In the previous theory, we have learnt how to find solutions using the graphical method.

Now, let us see how to find the solution to linear equations using the conditions.

Let us consider the system of linear equations in two variables.

Now, let us see how to find the solution to linear equations using the conditions.

Let us consider the system of linear equations in two variables.

\(a_1 x + b_1 y + c_1 = 0\)

\(a_2 x + b_2 y + c_2 = 0\)

Here, \(a_1\), \(a_2\), \(b_1\), \(b_2\), \(c_1\) and \(c_2\) are real numbers, and \(x\) and \(y\) are variables.

Now, let us see the condition for the system to be consistent and inconsistent.

**1**. If \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\), then the system of equations has a unique solution. Hence, the given system of equations is consistent. Furthermore, in the graphical representation, if the equations are consistent, the lines intersect at only one point.

**2**. If \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\), then the system of equations has infinitely many solutions. Hence, the given system of equations is consistent. In the graphical representation, if the equations are consistent, then the lines coincide with each other.

**3**. If \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\), then the system of equations has no solution. Hence, the given system of equations is inconsistent. If the equations are inconsistent in the graphical representation, then the lines are parallel to each other.

Let us verify the above theory with examples.

Example:

**1**. Check whether the pair of equations \(2x + y = 8\) and \(y = 2x\) are consistent or inconsistent.

**Solution**:

The pair of equations are:

\(2x + y = 8\) ---- (\(1\))

\(y = 2x \Rightarrow -2x + y = 0\) ---- (\(2\))

Compare the system of equations with the general equation.

\(a_1 x + b_1 y + c_1 = 0\)

\(a_2 x + b_2 y + c_2 = 0\)

Thus, we have:

\(a_1 = 2\), \(a_2 = -2\), \(b_1 = 1\), \(b_2 = 1\), \(c_1 = -8\), \(c_2 = 0\).

Now, let us compare the ratios.

\(\frac{a_1}{a_2} = \frac{2}{-2} = -1\)

\(\frac{b_1}{b_2} = \frac{1}{1} = 1\)

Since, \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\), then the system is consistent and has a unique solution.