Before solving linear equations, let us understand what are linear equations?

**Solving Linear Equations- In two variables**

Equations of degree one and having two different variables in it are called linear equations in two variables.

Standard form of a linear equation in two variables is,

\(ax~+~by~+~c\) = \(0\)

Where, \(a, ~b\) and \(c\) are real numbers, and both \(a\) and \(b\) are not equal to zero.

Equations of the form ax+b=0, are also linear equations in two variables; because \(ax~+~b\) = \(0\) can be written as \(ax~+~0~×~y~+~b\) = \(0\).

**Solution of linear equations**

\(2x~-~4\) = \(2\) is a linear equation in one variable. Every linear equation in one variable has a unique solution. While solving linear equations in one variable,

We can add same number on both sides of the equations.

\(\Rightarrow~2x~-~4~+~4\) = \(2~+~4\)

\(\Rightarrow~2x\) = \(6\)

We can divide both sides of the equation by same non-zero number. \(\Rightarrow~\frac{2x}{2}\) = \(\frac{6}{2}\)

\(\Rightarrow~x\) = \(3\)

Therefore, solution or root of the equation \(2x~-~4\) = \(2\) is \(3\).

**How to solve linear equations** in two variables is discussed below.

Since there are two different variables involved, a solution of linear equation in two variables means a pair of numbers. One for \(x\) and one for \(y\) which satisfy the given equation.

Consider the equation,

\(x~+~2y\) = \(10\)

Substitute \(x\) = \(2\) and \(y\) = \(4\),

\(\Rightarrow~LHS\) = \(2~+~8\) = \(10\) = \(RHS\)

\(x\) = \(2\) and \(y\) = \(4\) is a solution of this equation since it satisfy the equation.

This solution is written as ordered pair (\(2,4\)).

Similarly,

Substitute \(x\) = \(4\) and \(y\) = \(3\),

\(\Rightarrow~LHS\) = \(4~+~6\) = \(10\) = \(RHS\)

Therefore, (\(4,3\)) is also a solution of the equation.

Substitute \(x\) = \(3\) and \(y\) = \(4\),

\(\Rightarrow~LHS\) = \(3~+~8\) = \(11\) ≠ \(RHS\)

Therefore, (\(3,4\)) is not a solution of the equation.

We found that (\(2,4\)) and (\(4,3\)) are two solutions of the equation, \(x~+~2y\) = \(10\).

Similarly, (\(10,0\)) is another solution, since it satisfies the equation.

Many more solutions can be found for the above equation by the following method:

Fix a value for \(x\). For example \(x\) = \(1\), equation becomes,

\(1~+~2y\) = \(10\),

\(\Rightarrow~y\) = \(\frac{9}{2}\)

Therefore (\(1,~\frac{9}{2}\)) is another solution.

Similarly, fix \(x\) = \(6\),

\(\Rightarrow~6~+~2y\) = \(10\),

\(\Rightarrow~y\) = \(\frac{4}{2}\) = \(2\)

Therefore, (\(6,2\)) is also a solution of the equation.

Doing the same procedure with many numbers, many solutions can be found for the equation.

Therefore, a linear equation in two variables will have infinitely many solutions.

- Easy way to find two solutions of linear equations in two variables is, substituting \(x\) = \(0\) and find the corresponding value of \(y\). Similarly, substitute \(y\) = \(0\) and find the corresponding value of \(x\).

Example: Find two solutions of

- \(5x~+~4y\) = \(30\)

Put \(x\) = \(0\),

\(\Rightarrow~0~+~4y\) = \(30\)

\(\Rightarrow~y\) = \(\frac{30}{4}\) = \(\frac{15}{2}\)

(\(0,~\frac{15}{2}\))is a solution.

Put \(y\) = \(0\),

\(\Rightarrow~5x~+~0\) = \(30\)

\(\Rightarrow~x\) = \(\frac{30}{5}\) = \(6\)

(\(6,~0\)) is also a solution of \(5x~+~4y\) = \(30\)

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