# Solving Linear Equations in one Variable by Inverse Operations

The linear equations in one variable can be solved mathematically in a systematic method by the inverse operations. In this method, both sides of the linear equation is balanced by the basic mathematical operations inversely for making variable to appear at one side and its equivalent quantity to appear at the other side of the equation.

The inverse operations method is always recommendable and better than the Trial and Error method. In this method, four basic mathematical operations are used to separate the variable from the number in the linear equation in one variable.

If a variable is connected to a number by addition at one side of the equation, then use opposite operation subtraction in both sides of the linear equation for eliminating the number from one side of the equation completely.

$x+9 = 14$

In the left-hand side expression, the number $9$ is added to variable $x$. It should be eliminated from this expression to find the value of $x$. So, subtract $9$ from the left-hand side expression but it makes the right-hand side expression to become imbalanced. Therefore, subtract both sides of the equation by $9$ for making the equation systematic and balanced.

$\implies$ $x+9-9 = 14-9$

$\implies$ $\require{cancel} x+\cancel{9}-\cancel{9} = 14-9$

$\,\,\, \therefore \,\,\,\,\,\,$ $x = 5$

### Subtraction form

If a variable is connected to a number by subtraction at one side of the equation, then use its inverse operation addition at both sides of the linear equation for removing the number completely from one side of the equation.

$x-5 = 11$

The number $5$ is subtracted from variable $x$. In order to solve $x$, it must be eliminated from left-hand side expression and it is possible by adding $5$ to the expression but the expression in the right-hand side become imbalanced due to this operation. However, It can be balanced by adding $5$ to both sides of the equation.

$\implies$ $x-5+5 = 11+5$

$\implies$ $\require{cancel} x-\cancel{5}+\cancel{5} = 16$

$\,\,\, \therefore \,\,\,\,\,\,$ $x = 16$

### Multiplication form

If a variable is connected to a number by multiplication at one side the linear equation, then try its inverse operation division at both left and right-hand sides of the equation for solving the variable.

$4x = 24$

In this example, the number $4$ is multiplied to variable $x$. For finding the value of $x$, it’s essential to eliminate the number $4$ from the expression. It is usually done by the division but it imbalances the right-hand side of the equation. So, divide both sides the equation by the coefficient of the variable.

$\implies$ $\dfrac{4x}{4} = \dfrac{24}{4}$

$\implies$ $\require{cancel} \dfrac{\cancel{4}x}{\cancel{4}} = \dfrac{\cancel{24}}{\cancel{4}}$

$\,\,\, \therefore \,\,\,\,\,\,$ $x = 6$

### Division form

If a variable is connected to a number by division in one side the linear equation, then apply its inverse operation multiplication in both sides of the mathematical equation for evaluating the variable.

$\dfrac{x}{8} = 2$

The number $8$ divides the variable $x$ in this example. To solve the value of $x$, it is must to eliminate the number $8$ from the expression. It can be done by the multiplication and it imbalances the right-hand side expression. Therefore, it is essential to multiply both sides the equation by $8$.

$\implies$ $8 \times \dfrac{x}{8} = 8 \times 2$

$\implies$ $\dfrac{8x}{8} = 16$

$\implies$ $\require{cancel} \dfrac{\cancel{8}x}{\cancel{8}} = 16$

$\,\,\, \therefore \,\,\,\,\,\,$ $x = 16$

Thus, the inverse operations method is used in the linear equations in one variable for finding the unknown variable systematically.

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