
Solve the following equation: \[\dfrac{{5x - 7}}{{3x + 1}} = 1\].
Answer
601.2k+ views
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.
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