Find the solution for each inequality and graph it on a number line.
Final solutions:
The problem asks for the solutions to four separate inequality expressions. We analyze each one individually:
Each inequality will be examined for validity and solved for the variable \(x\) where applicable.
The first expression, \(5 \leq 24\), is a comparison between two numerical constants. Since 5 is indeed less than 24, this statement is universally true for any value of \(x\) that might have been implied, or simply serves as a true numerical assertion. No variable is involved here.
To solve \(\frac{x}{9} - 25 \leq -23\), we isolate \(x\):
First, add 25 to both sides of the inequality:
\(\frac{x}{9} \leq -23 + 25\)
\(\frac{x}{9} \leq 2\)
Next, multiply both sides by 9:
\(x \leq 2 \times 9\)
\(x \leq 18\)
The solution set is all real numbers \(x\) such that \(x \leq 18\). On a number line, this is represented by a solid circle at 18 with shading extending to the left.
The third expression, \(\leq 12\), is incomplete as it lacks a variable expression on the left-hand side. Without an expression involving \(x\), it cannot be solved or analyzed as a mathematical inequality. It is considered undefined or malformed.
To solve \(8x - 2\frac{1}{2} \geq 5\frac{1}{2}\), we first convert the mixed numbers to improper fractions:
\(2\frac{1}{2} = \frac{4+1}{2} = \frac{5}{2}\)
\(5\frac{1}{2} = \frac{10+1}{2} = \frac{11}{2}\)
The inequality becomes:
\(8x - \frac{5}{2} \geq \frac{11}{2}\)
Add \(\frac{5}{2}\) to both sides:
\(8x \geq \frac{11}{2} + \frac{5}{2}\)
\(8x \geq \frac{16}{2}\)
\(8x \geq 8\)
Divide by 8:
\(x \geq 1\)
The solution set is all real numbers \(x\) such that \(x \geq 1\). On a number line, this is represented by a solid circle at 1 with shading extending to the right.
Inequalities are fundamental in real-life resource management. For example, if a cargo elevator has a weight limit of 2,000 kg, and you are loading crates of 50 kg each, the number of crates \(c\) you can load satisfies the inequality \(50c \leq 2000\), which simplifies to \(c \leq 40\). Similarly, determining minimum requirements for projects or construction, such as requiring at least 100 meters of fencing for a garden, can be modeled as \(L \geq 100\). Mastering these operations allows you to set safe and effective limits in various practical tasks.