Question

1. $8 - x \geq 5(8 - x)$

1. $5 - x < 2(x - 3) + 5$

1. $\frac{x}{2} + 1 \leq 3x + 2$

1. $0.5x + 3 \geq 2x - 1.5$

Explanation:

Problem 11: \( 8 - x \geq 5(8 - x) \)

Step 1: Expand the right side

Expand \( 5(8 - x) \) to get \( 40 - 5x \). The inequality becomes \( 8 - x \geq 40 - 5x \).

Step 2: Add \( 5x \) to both sides

Adding \( 5x \) to both sides: \( 8 - x + 5x \geq 40 - 5x + 5x \), which simplifies to \( 8 + 4x \geq 40 \).

Step 3: Subtract 8 from both sides

Subtract 8: \( 8 + 4x - 8 \geq 40 - 8 \), so \( 4x \geq 32 \).

Step 4: Divide by 4

Divide both sides by 4: \( \frac{4x}{4} \geq \frac{32}{4} \), giving \( x \geq 8 \).

Step 1: Expand the right side

Expand \( 2(x - 3) + 5 \): \( 2x - 6 + 5 = 2x - 1 \). The inequality is \( 5 - x < 2x - 1 \).

Step 2: Add \( x \) to both sides

Adding \( x \): \( 5 - x + x < 2x - 1 + x \), so \( 5 < 3x - 1 \).

Step 3: Add 1 to both sides

Add 1: \( 5 + 1 < 3x - 1 + 1 \), giving \( 6 < 3x \).

Step 4: Divide by 3

Divide by 3: \( \frac{6}{3} < \frac{3x}{3} \), so \( 2 < x \) or \( x > 2 \).

Step 1: Subtract \( \frac{x}{2} \) from both sides

Subtract \( \frac{x}{2} \): \( 1 \leq 3x - \frac{x}{2} + 2 \). Simplify \( 3x - \frac{x}{2} = \frac{6x - x}{2} = \frac{5x}{2} \), so \( 1 \leq \frac{5x}{2} + 2 \).

Step 2: Subtract 2 from both sides

Subtract 2: \( 1 - 2 \leq \frac{5x}{2} \), so \( -1 \leq \frac{5x}{2} \).

Step 3: Multiply by \( \frac{2}{5} \)

Multiply both sides by \( \frac{2}{5} \): \( -1 \times \frac{2}{5} \leq x \), giving \( x \geq -\frac{2}{5} \) (or \( x \geq -0.4 \)).

Answer:

\( x \geq 8 \) (On the number line, a closed circle at 8 and an arrow to the right)

Problem 12: \( 5 - x < 2(x - 3) + 5 \)