Solve.
$5-\frac{1}{6} x=\frac{13}{3}$
We need to solve the linear equation:
\[5 - \frac{1}{6}x = \frac{13}{3}\]
Our variable is \(x\), and we aim to isolate it to find its value. The equation involves basic arithmetic operations with fractions and constants. Our plan is to first isolate the term containing \(x\), then simplify the constants, and finally solve for \(x\) by clearing the coefficient.
To isolate the term with \(x\), subtract 5 from both sides of the equation:
\[-\frac{1}{6}x = \frac{13}{3} - 5\]
Now, simplify the right side by expressing 5 as a fraction with a denominator of 3:
\[5 = \frac{5 \times 3}{3} = \frac{15}{3}\]
Substituting this back into the equation, we get:
\[-\frac{1}{6}x = \frac{13}{3} - \frac{15}{3}\]
\[-\frac{1}{6}x = -\frac{2}{3}\]
To solve for \(x\), multiply both sides of the equation by \(-6\):
\[x = \left(-\frac{2}{3}\right) \times (-6)\]
\[x = \frac{-2 \times -6}{3}\]
\[x = \frac{12}{3}\]
\[x = 4\]
Thus, the solution to the equation is \(4\).
In real-life scenarios, solving linear equations like this is essential for budgeting and resource management. For example, if you have a total of \(\$5\) and you want to spend a certain amount daily on a subscription fee \(\frac{1}{6}x\) such that you are left with \(\frac{13}{3} \approx \$4.33\) at the end of the term, you can use this equation to determine how many units of that fee you have incurred. By setting up and solving such equations, you can balance your finances accurately and make informed decisions about your spending.