Other than a no solution​ set, use interval notation to express the solution set and then graph the solution set on a number line.
2(9-3x)≤4-4x

To solve the inequality \(2(9 - 3x) \leq 4 - 4x\), follow these steps:

### Step 1: Distribute and simplify
Distribute the \(2\) on the left side:
\[
2(9 - 3x) = 18 - 6x
\]
So the inequality becomes:
\[
18 - 6x \leq 4 - 4x
\]

### Step 2: Move all terms involving \(x\) to one side
Add \(6x\) to both sides to move the variable terms to one side:
\[
18 \leq 4 + 2x
\]

### Step 3: Isolate the variable \(x\)
Subtract \(4\) from both sides:
\[
14 \leq 2x
\]
Now, divide by \(2\) to solve for \(x\):
\[
7 \leq x
\]

This can also be written as:
\[
x \geq 7
\]

### Step 4: Express the solution in interval notation
The solution \(x \geq 7\) in interval notation is:
\[
[7, \infty)
\]

### Step 5: Graph the solution on a number line
- Draw a number line.
- Place a solid dot at \(7\) to indicate that \(7\) is included in the solution (since the inequality is \(\geq\)).
- Shade the number line to the right of \(7\) to represent all numbers greater than or equal to \(7\).

Would you like a detailed explanation on any of these steps or have any other questions?

Here are 5 related questions you might find interesting:
1. How would the solution change if the inequality were strict, such as \(2(9 - 3x) < 4 - 4x\)?
2. How do you solve an inequality involving absolute values?
3. Can you solve a quadratic inequality and represent its solution on a number line?
4. What is the meaning of an open interval compared to a closed interval in interval notation?
5. How would the graph change if the inequality involved \(x \leq 7\) instead?

**Tip:** When multiplying or dividing both sides of an inequality by a negative number, remember to reverse the direction of the inequality sign.

Learn how to solve the inequality 2(9 - 3x) ≤ 4 - 4x with a detailed step-by-step solution using interval notation and number line graphing. This problem covers key concepts in inequalities and linear equations. Suitable for students in Grades 9-12.

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