Solve $\dfrac{x-3}{5}-2 = -1$ by Inverse Operations Method

$\dfrac{x-3}{5}-2 = -1$ is an equation in terms of a variable $x$. In this algebraic equation, the power of the variable $x$ is one and no other variable is involved. So, it is a linear equation in one variable and it can be solved systematically by the inverse operations method.

Eliminate -2 from expression by Addition

$\dfrac{x-3}{5}-2 = -1$

In left-hand side of the equation, the two terms are connected by subtraction. It is essential to eliminate $-2$ from this expression and it can be done by adding $2$ to both sides of the equation.

$\implies$ $\dfrac{x-3}{5}-2+2$ $=$ $-1+2$

$\implies$ $\require{cancel} \dfrac{x-3}{5}-\cancel{2}+\cancel{2}$ $=$ $1$

$\implies$ $\dfrac{x-3}{5} = 1$

Eliminate 5 from expression by Multiplication

$\dfrac{x-3}{5} = 1$

The number $5$ is dividing the algebraic expression $x-3$ in the left-hand side of the equation. It can be eliminated from this expression by multiplying both sides of the equation by $5$.

$\implies$ $5 \times \Bigg[\dfrac{x-3}{5}\Bigg]$ $=$ $5 \times 1$

$\implies$ $\dfrac{5(x-3)}{5}$ $=$ $5$

$\implies$ $\require{cancel} \dfrac{\cancel{5}(x-3)}{\cancel{5}}$ $=$ $5$

$\implies$ $x-3 = 5$

Eliminate -3 from expression by Addition

$x-3 = 5$

Two terms are connected by subtraction in the left-hand side of the equation. For solving $x$, the number $-3$ should be eliminated from this expression. It can be done by adding $3$ to both sides of the linear equation.

$\implies$ $x-3+3 = 5+3$

$\implies$ $x-3+3 = 8$

$\,\,\, \therefore \,\,\,\,\,\,$ $x = 8$

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