Question

Solve for x and graph the solution on the number line below, the graph goes from -12 to 12. -3→x/3

Answers

Answer
So x = -9?
Answer
Oh ok
Answered by GPT-4o mini
To solve the inequality \(-3 \to \frac{x}{3}\), we will interpret the \(\to\) symbol as representing the "less than" symbol (i.e., \(-3 < \frac{x}{3}\)).

Let’s solve this inequality step-by-step:

1. Start with the inequality:
\[
-3 < \frac{x}{3}
\]

2. To eliminate the fraction, multiply both sides by \(3\) (note that multiplying by a positive number does not change the direction of the inequality):
\[
-3 \cdot 3 < x
\]
\[
-9 < x
\]

3. We can also rewrite this inequality as:
\[
x > -9
\]

Now, we will graph the solution \(x > -9\) on a number line from -12 to 12.

### Number Line Graph
1. The value \(-9\) will be represented on the number line.
2. Since this is a strict inequality (>), we will use an open circle at \(-9\) (indicating that \(-9\) is not included in the solution).
3. Shade or draw an arrow to the right from \(-9\) to indicate all numbers greater than \(-9\).

Here’s a textual representation of how the graph should look:

```
-12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12
| | | (o)----------------------------->
```

In this representation:
- The open circle at \(-9\) indicates that \(-9\) is not included in the solution.
- The arrow points to the right, indicating that all numbers greater than \(-9\) are part of the solution set.
Answered by GPT-4o mini
No, \(x = -9\) is not the solution; rather, we found that \(x > -9\). This means that \(x\) can be any number greater than \(-9\).

The solution set is all real numbers greater than \(-9\). In interval notation, this is expressed as \((-9, \infty)\).

So, to clarify:
- The solution is not just \(x = -9\).
- The solution is \(x > -9\), which means any value larger than \(-9\) is acceptable.
Answered by GPT-4o mini
No problem! If you have any more questions or if there's anything else you'd like to know about solving inequalities or any other topic, feel free to ask!

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