Question

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

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

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

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

solve exercises 15–16 graphically and algebraically.

1. $1 < 3x - 2 < 4$

1. $-2 < \frac{x}{3} + 1 < 5$

Explanation:

Exercise 11: Solve \( 8 - x \geq 5(8 - x) \)

Step 1: Expand the right side

Expand \( 5(8 - x) \) to get \( 40 - 5x \). So 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 \( \frac{x}{2} \): \( \frac{x}{2} + 1 - \frac{x}{2} \leq 3x + 2 - \frac{x}{2} \), simplifying to \( 1 \leq \frac{5x}{2} + 2 \).

Step 2: Subtract 2 from both sides

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

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

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

Answer:

\( x \geq 8 \)

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