# Solve $\dfrac{6x-2}{9}$ $+$ $\dfrac{3x+5}{18}$ $=$ $\dfrac{1}{3}$

$\dfrac{6x-2}{9}$ $+$ $\dfrac{3x+5}{18}$ $=$ $\dfrac{1}{3}$ is a given algebraic equation. It is actually defined only in one variable $x$ and its exponent is one. Therefore, the equation in algebraic form is known as linear equation in one variable.

### Simplify the Linear Equation

$\dfrac{6x-2}{9}$ $+$ $\dfrac{3x+5}{18}$ $=$ $\dfrac{1}{3}$

In order to isolate $x$, the linear equation in one variable should be simplified and then it can be solved by using transposing method.

$\implies$ $\dfrac{6x}{9}$ $-$ $\dfrac{2}{9}$ $+$ $\dfrac{3x}{18}$ $+$ $\dfrac{5}{18}$ $=$ $\dfrac{1}{3}$

$\implies$ $\require{cancel} \dfrac{\cancel{6}x}{\cancel{9}}$ $-$ $\dfrac{2}{9}$ $+$ $\dfrac{\cancel{3}x}{\cancel{18}}$ $+$ $\dfrac{5}{18}$ $=$ $\dfrac{1}{3}$

$\implies$ $\dfrac{2x}{3}$ $-$ $\dfrac{2}{9}$ $+$ $\dfrac{x}{6}$ $+$ $\dfrac{5}{18}$ $=$ $\dfrac{1}{3}$

$\implies$ $\dfrac{2x}{3}$ $+$ $\dfrac{x}{6}$ $-$ $\dfrac{2}{9}$ $+$ $\dfrac{5}{18}$ $=$ $\dfrac{1}{3}$

$\implies$ $\dfrac{2 \times 2x + 1 \times x}{6}$ $+$ $\dfrac{5}{18}$ $-$ $\dfrac{2}{9}$ $=$ $\dfrac{1}{3}$

$\implies$ $\dfrac{4x+x}{6}$ $+$ $\dfrac{5}{18}$ $-$ $\dfrac{2}{9}$ $=$ $\dfrac{1}{3}$

$\implies$ $\dfrac{5x}{6}$ $+$ $\dfrac{1 \times 5 -2 \times 2}{18}$ $=$ $\dfrac{1}{3}$

$\implies$ $\dfrac{5x}{6}$ $+$ $\dfrac{5-4}{18}$ $=$ $\dfrac{1}{3}$

$\implies$ $\dfrac{5x}{6}$ $+$ $\dfrac{1}{18}$ $=$ $\dfrac{1}{3}$

### Use Transposing Method

$\dfrac{5x}{6}$ $+$ $\dfrac{1}{18}$ $=$ $\dfrac{1}{3}$

The linear equation in one variable is successfully simplified. Now, the transpose method can be applied for transposing the terms and it helps us to find the unknown.

In this case, the term $\dfrac{1}{18}$ should be transposed to right-hand side of the equation. It is a first step for isolating the $x$ in this linear equation problem.

$\implies$ $\dfrac{5x}{6}$ $=$ $\dfrac{1}{3}$ $-$ $\dfrac{1}{18}$

$\implies$ $\dfrac{5x}{6}$ $=$ $\dfrac{6 \times 1 -1 \times 1}{18}$

$\implies$ $\dfrac{5x}{6}$ $=$ $\dfrac{6-1}{18}$

$\implies$ $\dfrac{5x}{6}$ $=$ $\dfrac{5}{18}$

$\implies$ $6 \times \dfrac{5x}{6}$ $=$ $6 \times \dfrac{5}{18}$

$\implies$ $\dfrac{6 \times 5x}{6}$ $=$ $\dfrac{6 \times 5}{18}$

$\implies$ $\require{cancel} \dfrac{\cancel{6} \times 5x}{\cancel{6}}$ $=$ $\dfrac{\cancel{6} \times 5}{\cancel{18}}$

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

### Solve the unknown of equation

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

The role of transposing method is completed. Now, the root of the linear equation in one variable can be evaluated by the inverse operations method.

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

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

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

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

$\implies$ $x$ $=$ $\require{cancel} \dfrac{\cancel{5} \times 1}{3 \times \cancel{5}}$

$\,\,\, \therefore \,\,\,\,\,\,$ $x$ $=$ $\dfrac{1}{3}$

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