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Hint: If we solve the given equation it converts into a linear equation in one variable. In the given equation the variable is \[x\]. Since it is a linear equation in one variable, if we solve the equation, we will get one only one solution for the variable \[x\]. So, use this concept to reach the solution of the problem.

Complete step-by-step answer:

Given equation is \[\dfrac{{5x - 7}}{{3x + 1}} = 1\]

Taking the denominator to the R.H.S, we have

\[

5x - 7 = 1\left( {3x + 1} \right) \\

5x - 7 = 3x + 1 \\

\]

Grouping the variable terms, we get

\[5x - 3x = 1 + 7\]

Taking the variable as common, we have

\[

\left( {5 - 3} \right)x = 1 + 7 \\

2x = 8 \\

\]

Therefore, we have

\[

x = \dfrac{8}{2} = 4 \\

\therefore x = 4 \\

\]

Thus, the value of the variable \[x\]in equation \[\dfrac{{5x - 7}}{{3x + 1}} = 1\]is 4.

Note: This problem involves solving a linear equation in one variable with variables on both sides by taking the denominator to the right-hand side (R.H.S). The degree (highest power) of the linear equation is one. So, we have only one solution to the equation.

Complete step-by-step answer:

Given equation is \[\dfrac{{5x - 7}}{{3x + 1}} = 1\]

Taking the denominator to the R.H.S, we have

\[

5x - 7 = 1\left( {3x + 1} \right) \\

5x - 7 = 3x + 1 \\

\]

Grouping the variable terms, we get

\[5x - 3x = 1 + 7\]

Taking the variable as common, we have

\[

\left( {5 - 3} \right)x = 1 + 7 \\

2x = 8 \\

\]

Therefore, we have

\[

x = \dfrac{8}{2} = 4 \\

\therefore x = 4 \\

\]

Thus, the value of the variable \[x\]in equation \[\dfrac{{5x - 7}}{{3x + 1}} = 1\]is 4.

Note: This problem involves solving a linear equation in one variable with variables on both sides by taking the denominator to the right-hand side (R.H.S). The degree (highest power) of the linear equation is one. So, we have only one solution to the equation.