Equations can be classified into linear, quadratic or cubic etc. Similarly linear equations can be classified into those containing, one variable, two variables or three or more variables.

While solving these equations we follow the rules of equality we discussed in earlier sections. To solve the linear equation of only one variable, it is enough if we have only one equation. While solving linear equations containing two variables we need two equations to solve for the unknown variables.

In this section let us see how to solve equations with variables on both sides and discuss with few examples also.

## How to Solve Equations with Variables on Both Sides

Solution to a given equation is the value of the variable, which satisfy the equations.

Rules to solve equations with variables on both sides:

Linear Equations: The general form of linear equations will be ax + b = 0, where a $\ne$ 0.
In the case of linear equations which have only one variable for which the variables are on both sides, we need to bring the variables to one side of the equal to sign, and the constants to other side of the equation and then eliminate the constants with the variable and solve for x.

### Example: 2x - 5 = 13 - 4x

Solution: We have 2x - 5 = 13 - 4x
Eliminating the constant in the left hand side, we get,
2x - 5 + 5 = 13 - 4x + 5
=> 2x = 18 - 4x
=> 2x + 4x = 18 - 4x + 4x
=> 6x = 18

=> x = $\frac{18}{6}$

= 3
Therefore, solution x = 3

Quadratic Equations: The quadratic equations are of the form, ax2 + b x + c = 0, where a $\ne$ 0.
A quadratic equation is a polynomial of degree two. There will be two solutions for the quadratic equations. The solutions are called as the roots of the equations.
The equations can be solved by
a. Factorization
b. Formula Method
c. Completing the square method.

When the equations have variables on both sides, we should follow the rules of equality, such that all the terms containing the variables and the constants are brought to the left side of the equation, so that the final equation is of the form, ax2 + bx + c = 0, then we solve the equation by one of the three methods shown above.

### Example: Solve( x + 2 ) = $\frac{(6x+3)}{(x-2)}$

Solution:
We have ( x + 2 ) = $\frac{(6x+3)}{(x-2)}$

( x + 2 ) ( x - 2 ) = 6x + 3 [ by multiplying by ( x - 2 ) on both sides ]
x2 - 4 = 6x + 3
=> x2 - 6x - 4 - 3 = 0
=> x2 - 6x - 7 = 0
=> x2 - 7x + x - 7 = 0
=> x ( x - 7 ) + 1 ( x - 7 ) = 0
=> ( x - 7 ) ( x + 1 ) = 0
=> x - 7 = 0 or x + 1 = 0
=> x = 7 or x = -1
Therefore the solution to the above equation is x = { -1, 7}.

## Examples of Solving Equations with Variables on Both Sides

### Solved Examples

Question 1: Solve 3x - $\frac{x}{2}$ = $\frac{3}{2}$
Solution:

We have 3x - $\frac{x}{2}$ = $\frac{3}{2}$

In the denominator we have only 2.

Multiplying both sides of the equation by 2, we get

3x $\times$ 2 - $\frac{x}{2}$ $\times$ 2 = $\frac{3}{2}$ $\times$ 2

=> 6x - x = 3

=> 5x = 3 [ applying the rule of equality for elimination on either sides of the equation ]

=>                            x = $\frac{3}{5}$ is the solution to the above equation.

Question 2: Solve 2 + $\frac{(2x-3)}{6}$ = $\frac{(3x+4)}{5}$
Solution:

We have 2 + $\frac{(2x-3)}{6}$ = $\frac{(3x+4)}{5}$

=>  2 x 30 + $\frac{(2x-3)}{6}$ $\times$ 30 = $\frac{(3x+4)}{5}$ $\times$ 30 [ Multiplying by the LCM of the denominators 6 and 5 which is 30 ]

=>               60 + 5 ( 2x - 3 ) = 6 ( 3x + 4 )

=>                 60 + 10 x - 15 = 18 x + 24  [ applying the rule of equality for elimination on either sides of the equation ]

=>                      10 x - 18 x = 24 - 60 + 15

=>                                - 8x = -21

=>                                   x = $\frac{-21}{-8}$

= $\frac{21}{8}$

x = $\frac{21}{8}$ is the solution to the above equation.

Question 3: Solve $\frac{100}{x}$ - $\frac{100}{(x+5)}$ = 1
Solution:

The Denominators are x and ( x + 5 ).
LCM of the denominators = x ( x + 5 )

Multiplying by x ( x + 5 ) on both sides of the equation, we get,

$\frac{100}{x}$ $\times$ x ( x + 5 )  - $\frac{100}{(x+5)}$ $\times$ x ( x + 5 ) = 1 $\times$ x ( x + 5 )

100 ( x + 5 ) - 100 x = x ( x + 5 )
=>          100 x + 500 - 100x = x2 + 5 x
=>                 x2 + 5 x - 500 = 0 [ applying the rule of equality for elimination on either sides of the equation ]
=>      x2 + 25 x - 20 x - 500 = 0
=>  x ( x + 25 ) - 20 ( x + 25 ) = 0
=>           ( x - 20 ) ( x + 25 ) = 0
=>                             x - 20 = 0  or  x + 25 = 0
=>                                   x = 20   or  x = - 25
The solution will be x = { 20, - 25 }.