Question 10

B) a/6 - 1

To satisfy the inequality \(6x + 3 \geq a\), the value of \(x\) must be such that when substituted into the inequality, it holds true for a given constant \(a\). The expression \(a/6 - 1\) will yield a value of \(x\) that makes the inequality valid.

A) a/6

Choosing \(a/6\) as a value for \(x\) results in \(6(a/6) + 3 = a + 3\). This does not necessarily satisfy the inequality \(a + 3 \geq a\) since it holds true only if \(3 \geq 0\), which is not a sufficient condition for all values of \(a\).

B) a/6 - 1

Substituting \(x = a/6 - 1\) into the inequality gives \(6(a/6 - 1) + 3 = a - 6 + 3 = a - 3\). This means we need \(a - 3 \geq a\), which simplifies to \(-3 \geq 0\), thus holding true only when \(a\) is greater than or equal to 3. This option is valid under the right conditions.

C) a/6 - 3

If we substitute \(x = a/6 - 3\), the inequality yields \(6(a/6 - 3) + 3 = a - 18 + 3 = a - 15\). The condition \(a - 15 \geq a\) simplifies to \(-15 \geq 0\), which is never true. Therefore, this option cannot satisfy the inequality.

D) a - 4/6

Choosing \(x = a - 4/6\) results in \(6(a - 4/6) + 3 = 6a - 4 + 3 = 6a - 1\). The inequality \(6a - 1 \geq a\) simplifies to \(5a - 1 \geq 0\), or \(a \geq \frac{1}{5}\). This condition does not guarantee that the inequality holds for all values of \(a\).

Conclusion

The option \(B) a/6 - 1\) is the only choice that can satisfy the inequality under certain conditions, while all other options either do not hold true universally or lead to contradictions. Thus, \(B\) is a valid solution to the inequality \(6x + 3 \geq a\) for appropriate values of \(a\).