Solve the inequality $f+11 \leq \frac{5}{3}$ and write the solution in interval notation. Simplify your answer and enter as a fraction.

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04-11-2026
Mathematics

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Solve the inequality $f+11 \leq \frac{5}{3}$ and write the solution in interval notation. Simplify your answer and enter as a fraction.

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GinnyAnswer GinnyAnswer · Answer 1 · 2026-11-04T00:06:17-05:00

Subtract 11 from both sides of the inequality: $f \leq \frac{5}{3} - 11$.
Convert 11 to a fraction with a common denominator: $11 = \frac{33}{3}$.
Simplify the right side: $\frac{5}{3} - \frac{33}{3} = -\frac{28}{3}$.
Express the final inequality in interval notation: $\boxed{(-\infty, -\frac{28}{3}]}$.

Explanation

Analysis of the Inequality

We are tasked with solving the linear inequality $f + 11 \leq \frac{5}{3}$ and expressing the solution in interval notation. The variable $f$ is isolated on the left side of the inequality. To solve for $f$, we need to isolate it by moving the constant term 11 to the right side. We do this by subtracting 11 from both sides of the inequality.

Isolating the Variable f

Starting with the inequality:
\[f + 11 \leq \frac{5}{3}\]
Subtract 11 from both sides:
\[f \leq \frac{5}{3} - 11\]
To perform the subtraction on the right side, we need a common denominator. Since the denominator of the first term is 3, we can rewrite 11 as a fraction with a denominator of 3:
\[11 = \frac{11 \times 3}{3} = \frac{33}{3}\]
Now, we substitute this back into the inequality:
\[f \leq \frac{5}{3} - \frac{33}{3}\]

Final Solution and Interval Notation

Now, perform the subtraction of the fractions on the right side:
\[f \leq \frac{5 - 33}{3}\]
\[f \leq -\frac{28}{3}\]
The inequality $f \leq -\frac{28}{3}$ represents all real numbers less than or equal to $-\frac{28}{3}$. In interval notation, this is expressed by including all values from $-\infty$ up to and including $-\frac{28}{3}$. The notation for this is $(-\infty, -\frac{28}{3}]$.

Examples

In everyday life, this type of calculation can be used in budgeting or project planning. Suppose you have a limit on the total weight of a package, and you know the weight of the box itself is 11 units. If the total capacity is limited to $\frac{5}{3}$ units, you can easily determine how much additional weight you can put in the package by solving $f + 11 \leq \frac{5}{3}$. This ensures that you stay within the capacity constraints.