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

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#### Practise This Question

The linear equation 2x – 5y = x – 2 has