Question

which of the following is a solution to the inequality below?
\\(\frac{h}{3} - 1 \leq 5\\)
\\(h = 126\\) \\(h = 42\\) \\(h = 99\\) \\(h = 3\\)

Explanation:

Step1: Solve the inequality for \( h \)

First, add 1 to both sides of the inequality \(\frac{h}{3}-1\leq5\).
\(\frac{h}{3}-1 + 1\leq5 + 1\)
Simplifying both sides, we get \(\frac{h}{3}\leq6\).

Step2: Multiply both sides by 3

Multiply both sides of the inequality \(\frac{h}{3}\leq6\) by 3 to isolate \( h \).
\(3\times\frac{h}{3}\leq6\times3\)
Simplifying, we find \(h\leq18\).

Step3: Check each option

• For \( h = 126 \): \(126>18\), so it is not a solution.
• For \( h = 42 \): \(42>18\), so it is not a solution.
• For \( h = 99 \): \(99>18\), so it is not a solution.
• For \( h = 3 \): \(3\leq18\), so it is a solution.

Answer:

\( h = 3 \)