# Solving Linear Equations

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

1. $$5x~+~4y$$ = $$30$$

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 $$5x~+~4y$$ = $$30$$<