Find the possible values for $s$ in the inequality $12 s-20 \leq 50-3 s-25$.
A) $s \leq 1 / 3$
B) $s \leq 9$
C) $s \leq 5 / 9$
D) $s \leq 3$
Answer (1)
Simplify the inequality: 12 s − 20 ≤ 50 − 3 s − 25 becomes 12 s − 20 ≤ 25 − 3 s .
Add 3 s to both sides: 15 s − 20 ≤ 25 .
Add 20 to both sides: 15 s ≤ 45 .
Divide by 15 : s ≤ 3 . The solution is s ≤ 3 .
Explanation
Understanding the Inequality We are given the inequality 12 s − 20 ≤ 50 − 3 s − 25 . Our goal is to isolate s on one side of the inequality to find the possible values it can take.
Simplifying the Right-Hand Side First, simplify the right-hand side of the inequality by combining the constants: 50 − 25 = 25 . So the inequality becomes 12 s − 20 ≤ 25 − 3 s .
Adding 3 s to Both Sides Next, add 3 s to both sides of the inequality to get the s terms on the same side: 12 s + 3 s − 20 ≤ 25 − 3 s + 3 s , which simplifies to 15 s − 20 ≤ 25 .
Adding 20 to Both Sides Now, add 20 to both sides of the inequality to isolate the s term: 15 s − 20 + 20 ≤ 25 + 20 , which simplifies to 15 s ≤ 45 .
Dividing by 15 Finally, divide both sides of the inequality by 15 to solve for s : 15 15 s ≤ 15 45 , which simplifies to s ≤ 3 .
Finding the Solution Therefore, the possible values for s are s ≤ 3 . Comparing this to the given options, we see that option D is the correct answer.
Examples
Imagine you're managing a budget for a school event. You have a total of $50 to spend, but you need to subtract $20 for fixed costs and $3 for each student attending. If you want to ensure you have at least $25 left for other expenses, this inequality helps you determine the maximum number of students you can accommodate. By solving the inequality, you find the limit on the number of students to stay within your budget while meeting your financial goals.
