Question

which graph shows the solution of $-6(n - 1) < 3$ and $2(n + 1) > 0$?

a)

b)

c)

d)

Explanation:

Step1: Solve first inequality

Solve \(-6(n - 1) < 3\). Divide both sides by \(-6\) (remember to reverse inequality sign): \(n - 1 > \frac{3}{-6}=-\frac{1}{2}\). Add 1 to both sides: \(n > -\frac{1}{2}+1=\frac{1}{2}=0.5\)? Wait, no: \(-\frac{1}{2}+1 = \frac{1}{2}\)? Wait, \(-\frac{1}{2}+1=\frac{1}{2}\)? Wait, \(1=\frac{2}{2}\), so \(\frac{2}{2}-\frac{1}{2}=\frac{1}{2}\)? No, wait, \(-6(n - 1) < 3\) → divide by -6: \(n - 1 > -\frac{3}{6}=-\frac{1}{2}\) → \(n > -\frac{1}{2}+1=\frac{1}{2}\)? Wait, no, \(1\) is \(\frac{2}{2}\), so \(-\frac{1}{2}+\frac{2}{2}=\frac{1}{2}\)? Wait, no, that's wrong. Wait, \(-6(n - 1) < 3\) → \(n - 1 > -\frac{3}{6}=-\frac{1}{2}\) → \(n > -\frac{1}{2}+1\)? Wait, no, adding 1 to both sides: \(n > -\frac{1}{2}+1\)? Wait, \(1\) is \(\frac{2}{2}\), so \(-\frac{1}{2}+\frac{2}{2}=\frac{1}{2}\)? Wait, no, that's incorrect. Wait, let's do it again: \(-6(n - 1) < 3\) → divide both sides by -6 (inequality flips): \(n - 1 > \frac{3}{-6}=-\frac{1}{2}\) → then \(n > -\frac{1}{2}+1\)? Wait, no, \(n - 1 > -\frac{1}{2}\) → add 1 to both sides: \(n > -\frac{1}{2}+1\)? Wait, \(1\) is \(\frac{2}{2}\), so \(-\frac{1}{2}+\frac{2}{2}=\frac{1}{2}\)? Wait, no, that's not right. Wait, \(-\frac{1}{2}+1 = \frac{1}{2}\)? Wait, \(1 - 0.5 = 0.5\), yes. Wait, but let's check the second inequality.

Step2: Solve second inequality

Solve \(2(n + 1) > 0\). Divide both sides by 2: \(n + 1 > 0\) → \(n > -1\).

Step3: Find intersection

We have \(n > -\frac{1}{2}\) (from first inequality: wait, no, wait first inequality: \(-6(n - 1) < 3\) → \(-6n + 6 < 3\) → \(-6n < -3\) → divide by -6: \(n > \frac{-3}{-6}=\frac{1}{2}=0.5\). Oh! I made a mistake earlier. Let's redo first inequality:

\(-6(n - 1) < 3\)

Expand: \(-6n + 6 < 3\)

Subtract 6: \(-6n < 3 - 6 = -3\)

Divide by -6 (reverse inequality): \(n > \frac{-3}{-6}=\frac{1}{2}=0.5\)

Second inequality: \(2(n + 1) > 0\) → \(n + 1 > 0\) → \(n > -1\)

So the solution to the system is the intersection of \(n > 0.5\) and \(n > -1\), which is \(n > 0.5\). Wait, no: wait, first inequality: \(n > 0.5\), second: \(n > -1\). The intersection is \(n > 0.5\) (since 0.5 > -1). Now check the graphs:

Graph A: two open circles, one at -1, one at 0.5, lines going left from -1 and right from 0.5? No, the arrows: A has left from -1 and right from 0.5.

Graph B: open circle at -1, line going right.

Graph C: open circle at 0.5, line going right.

Graph D: open circle at -1 and 0.5, line between them? Wait, no, let's look at the options:

Wait, the problem is to solve the system of inequalities: \(-6(n - 1) < 3\) and \(2(n + 1) > 0\).

First inequality solution: \(n > 0.5\)

Second inequality solution: \(n > -1\)

The solution to the system is \(n > 0.5\) (since it's the intersection of \(n > 0.5\) and \(n > -1\); the more restrictive is \(n > 0.5\)).

Now check the graphs:

Option C: open circle at 0.5, line going right. That matches \(n > 0.5\) (open circle because it's >, not ≥).

Wait, let's re-express the first inequality correctly:

\(-6(n - 1) < 3\)

Divide both sides by -6: \(n - 1 > \frac{3}{-6} = -\frac{1}{2}\) → \(n > -\frac{1}{2} + 1 = \frac{1}{2}\)? Wait, no, that's wrong. Wait, \(-6(n - 1) < 3\) → \(n - 1 > -\frac{3}{6}\) → \(n - 1 > -\frac{1}{2}\) → \(n > -\frac{1}{2} + 1 = \frac{1}{2}\)? Wait, no, \(1\) is \(\frac{2}{2}\), so \(-\frac{1}{2} + \frac{2}{2} = \frac{1}{2}\). Wait, but when we did the other method: \(-6n + 6 < 3\) → \(-6n < -3\) → \(n > \frac{-3}{-6} = \frac{1}{2}\). Yes, so \(n > 0.5\).

Second inequality: \(2(n + 1) > 0\) → \(n + 1 > 0\) → \(n > -1\).

So the solution is \(n > 0.5\) (since 0.5 > -1, the intersection is \(n > 0.5\)). Now look at the graphs:

- Graph A: two open circles (at -1 and 0.5), line left from -1, right from 0.5. Not correct.
- Graph B: open circle at -1, line right. Not correct (needs n > 0.5).
- Graph C: open circle at 0.5, line right. Correct, because \(n > 0.5\) (open circle, since it's >, not ≥).
- Graph D: open circles at -1 and 0.5, line between them. No, that's \( -1 < n < 0.5\), which is wrong.

Wait, but let's check again. Wait, first inequality: \(n > 0.5\), second: \(n > -1\). So the solution is \(n > 0.5\), because 0.5 is greater than -1, so the intersection is the set of numbers greater than 0.5. So the graph should have an open circle at 0.5 and an arrow to the right. That's graph C.

Wait, but let's confirm the first inequality again:

\(-6(n - 1) < 3\)

Divide both sides by -6 (reverse inequality):

\(n - 1 > \frac{3}{-6} = -\frac{1}{2}\)

Add 1 to both sides:

\(n > -\frac{1}{2} + 1 = \frac{1}{2} = 0.5\). Yes, that's correct.

Second inequality: \(2(n + 1) > 0\) → \(n + 1 > 0\) → \(n > -1\). So the solution is \(n > 0.5\) (since 0.5 > -1, the intersection is \(n > 0.5\)). So the graph with open circle at 0.5 and arrow right is option C.

Answer:

C. The graph with an open circle at 0.5 and an arrow pointing to the right (Option C)