Let's solve the system of inequalities \( 5x - 2 \leq x + 2 \) and \( 2x - 4 \geq x + 1 \) separately and then find the intersection of the solutions.

### Step 1: Solve the first inequality

\[
5x - 2 \leq x + 2
\]

Subtract \( x \) from both sides to get all terms involving \( x \) on one side:

\[
4x - 2 \leq 2
\]

Add 2 to both sides:

\[
4x \leq 4
\]

Divide by 4:

\[
x \leq 1
\]

### Step 2: Solve the second inequality

\[
2x - 4 \geq x + 1
\]

Subtract \( x \) from both sides:

\[
x - 4 \geq 1
\]

Add 4 to both sides:

\[
x \geq 5
\]

### Step 3: Find the intersection of the solutions

From the first inequality, we have \( x \leq 1 \), and from the second inequality, we have \( x \geq 5 \).

Since \( x \) cannot be simultaneously less than or equal to 1 and greater than or equal to 5, there is **no solution** that satisfies both inequalities.

Would you like more details or have any questions? Here are some related problems to try:

1. Solve the inequality \( 3x + 1 \leq x + 7 \).
2. Solve \( 4x - 3 \geq 2x + 5 \).
3. Determine the solution to \( 6x - 2 \leq 3x + 8 \).
4. Solve the system: \( x - 2 \geq 4 \) and \( 3x + 1 \leq 2x + 5 \).
5. Solve \( 5x + 3 \geq 6x - 2 \).

**Tip:** When solving systems of inequalities, always carefully consider the intersection of the solution sets. If the intervals don't overlap, there's no solution.

Learn how to solve the system of inequalities 5x-2 ≤ x+2 and 2x-4 ≥ x+1 with a detailed step-by-step solution. This problem covers key algebraic concepts related to inequalities and linear equations. Suitable for high school students.

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