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In the following, determine whether the given values are solutions of the given equation or not:
$x\ +\ \frac{1}{x}\ =\ \frac{13}{6},\ x\ =\ \frac{5}{6},\ x\ =\ \frac{4}{3}$
Given:
The given equation is $x\ +\ \frac{1}{x}\ =\ \frac{13}{6}$.
To do:
We have to determine whether $x=\frac{5}{6}, x=\frac{4}{3}$ are solutions of the given equation.
Solution:
If the given values are the solutions of the given equation then they should satisfy the given equation.
Therefore,
For $x=\frac{5}{6}$,
LHS$=x + \frac{1}{x}$.
$=\frac{5}{6}+\frac{1}{\frac{5}{6}}$
$=\frac{5}{6}+\frac{6}{5}$
$=\frac{5\times5+6\times6}{30}$ (LCM of $5$ and $6$ is $30$)
$=\frac{25+36}{30}$
$=\frac{61}{30}$
$≠$RHS
Hence, $x=\frac{5}{6}$ is not a solution of the given equation.
For $x=\frac{4}{3}$,
LHS$=\frac{4}{3} +\frac{1}{\frac{4}{3}}$
$=\frac{4}{3}+\frac{3}{4}$
$=\frac{4\times4+3\times3}{12}$ (LCM of $3$ and $4$ is $12$)
$=\frac{16+9}{12}$
$=\frac{25}{12}$
RHS$=\frac{13}{6}$
LHS$≠$RHS
Hence, $x=\frac{4}{3}$ is not a solution of the given equation.
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