Question

-3 < y - 4 < 1

6
-2x + 7 < 1 or
4x + 3 ≤ -5

7
2x - 7 > 1 or
4x + 3 ≤ 5

Explanation:

Problem 6: Solve \(-2x + 7 < 1\) or \(4x + 3 \leq -5\)

Step 1: Solve \(-2x + 7 < 1\)

Subtract 7 from both sides:
\(-2x + 7 - 7 < 1 - 7\)
\(-2x < -6\)

Divide both sides by \(-2\) (reverse inequality sign):
\(x > \frac{-6}{-2}\)
\(x > 3\)

Step 2: Solve \(4x + 3 \leq -5\)

Subtract 3 from both sides:
\(4x + 3 - 3 \leq -5 - 3\)
\(4x \leq -8\)

Divide both sides by 4:
\(x \leq \frac{-8}{4}\)
\(x \leq -2\)

Step 3: Combine solutions (OR inequality)

For "or" inequalities, the solution is the union of the two solution sets:
\(x > 3\) or \(x \leq -2\)

Step 1: Solve \(2x - 7 > 1\)

Add 7 to both sides:
\(2x - 7 + 7 > 1 + 7\)
\(2x > 8\)

Divide both sides by 2:
\(x > \frac{8}{2}\)
\(x > 4\)

Step 2: Solve \(4x + 3 \leq 5\)

Subtract 3 from both sides:
\(4x + 3 - 3 \leq 5 - 3\)
\(4x \leq 2\)

Divide both sides by 4:
\(x \leq \frac{2}{4}\)
\(x \leq \frac{1}{2}\)

Step 3: Combine solutions (OR inequality)

For "or" inequalities, the solution is the union of the two solution sets:
\(x > 4\) or \(x \leq \frac{1}{2}\)

Answer:

\(x \leq -2\) or \(x > 3\)

Problem 7: Solve \(2x - 7 > 1\) or \(4x + 3 \leq 5\)