Question

1. graph the compound inequali 3x + 1 < 4 or 2x - 5 > 7

Explanation:

Step1: Solve \(3x + 1 < 4\)

Subtract 1 from both sides: \(3x + 1 - 1 < 4 - 1\)
Simplify: \(3x < 3\)
Divide both sides by 3: \(\frac{3x}{3} < \frac{3}{3}\)
Result: \(x < 1\)

Step2: Solve \(2x - 5 > 7\)

Add 5 to both sides: \(2x - 5 + 5 > 7 + 5\)
Simplify: \(2x > 12\)
Divide both sides by 2: \(\frac{2x}{2} > \frac{12}{2}\)
Result: \(x > 6\)

Step3: Graph the solutions

• For \(x < 1\), draw an open circle at \(x = 1\) (since 1 is not included) and shade to the left.
• For \(x > 6\), draw an open circle at \(x = 6\) (since 6 is not included) and shade to the right.
• The "or" means we combine these two shaded regions.

Answer:

The solution to the compound inequality \(3x + 1 < 4\) or \(2x - 5 > 7\) is \(x < 1\) or \(x > 6\). The graph has an open circle at \(1\) with shading left, an open circle at \(6\) with shading right.