Questions: Topic: Compound Inequalities Progress: Which one of the following compound inequalities has no solution? 3x+4<x-8 and -5x+4>3x-10 4(x+1) ≥ 3(x+2) and -3(x-1)<5(x+2) -4x-3 ≤ 2x+9 and -3x+2>-2x+5 4x-6 ≤ 5x+6 and -3x-2>2x+8

Solution Steps

To determine which compound inequality has no solution, we need to solve each pair of inequalities separately and then check if there is any overlap in their solutions. If there is no overlap, then the compound inequality has no solution.

1. Solve each inequality in the pair separately.
2. Determine the solution set for each inequality.
3. Check if there is any overlap between the solution sets of the two inequalities in each pair.
4. Identify the pair with no overlapping solution sets.

The first compound inequality is: \[ 3x + 4 < x - 8 \] \[ -5x + 4 > 3x - 10 \]

1. Subtract \( x \) from both sides: \[ 3x + 4 - x < x - 8 - x \] \[ 2x + 4 < -8 \]
2. Subtract 4 from both sides: \[ 2x + 4 - 4 < -8 - 4 \] \[ 2x < -12 \]
3. Divide by 2: \[ x < -6 \]

1. Add \( 5x \) to both sides: \[ -5x + 4 + 5x > 3x - 10 + 5x \] \[ 4 > 8x - 10 \]
2. Add 10 to both sides: \[ 4 + 10 > 8x - 10 + 10 \] \[ 14 > 8x \]
3. Divide by 8: \[ \frac{14}{8} > x \] \[ \frac{7}{4} > x \] \[ x < \frac{7}{4} \]

\[ x < -6 \] and \[ x < \frac{7}{4} \]

Since \( x < -6 \) is more restrictive, the solution to the first compound inequality is: \[ x < -6 \]

The second compound inequality is: \[ 4(x + 1) \geq 3(x + 2) \] \[ -3(x - 1) < 5(x + 2) \]

1. Distribute: \[ 4x + 4 \geq 3x + 6 \]
2. Subtract \( 3x \) from both sides: \[ 4x + 4 - 3x \geq 3x + 6 - 3x \] \[ x + 4 \geq 6 \]
3. Subtract 4 from both sides: \[ x + 4 - 4 \geq 6 - 4 \] \[ x \geq 2 \]

1. Distribute: \[ -3x + 3 < 5x + 10 \]
2. Add \( 3x \) to both sides: \[ -3x + 3 + 3x < 5x + 10 + 3x \] \[ 3 < 8x + 10 \]
3. Subtract 10 from both sides: \[ 3 - 10 < 8x + 10 - 10 \] \[ -7 < 8x \]
4. Divide by 8: \[ \frac{-7}{8} < x \] \[ x > -\frac{7}{8} \]

\[ x \geq 2 \] and \[ x > -\frac{7}{8} \]

Since \( x \geq 2 \) is more restrictive, the solution to the second compound inequality is: \[ x \geq 2 \]

The third compound inequality is: \[ -4x - 3 \leq 2x + 9 \] \[ -3x + 2 > -2x + 5 \]

1. Add \( 4x \) to both sides: \[ -4x - 3 + 4x \leq 2x + 9 + 4x \] \[ -3 \leq 6x + 9 \]
2. Subtract 9 from both sides: \[ -3 - 9 \leq 6x + 9 - 9 \] \[ -12 \leq 6x \]
3. Divide by 6: \[ \frac{-12}{6} \leq x \] \[ -2 \leq x \] \[ x \geq -2 \]

1. Add \( 3x \) to both sides: \[ -3x + 2 + 3x > -2x + 5 + 3x \] \[ 2 > x + 5 \]
2. Subtract 5 from both sides: \[ 2 - 5 > x + 5 - 5 \] \[ -3 > x \] \[ x < -3 \]

\[ x \geq -2 \] and \[ x < -3 \]

Since there is no \( x \) that can satisfy both \( x \geq -2 \) and \( x < -3 \) simultaneously, the third compound inequality has no solution.

Final Answer