Question

solve. match each problem number with its solution graph let

1. $3x - 1 > 3 + x$
2. $\frac{1}{2}(x - 4) > x + 8$
3. $1 - 8x \geq 3x - 10$
4. $2x + 6 < 5x - 3$
5. $-4(x - 8) \geq -2(2x - 16)$
6. $3x - 2 \leq 2x + 1$
7. $-2(8 - 3x) > 6x + 2$
8. $-3(3x + 5) > -5(2x - 2)$
9. $8(2x - 1) > 11x + 17$
10. $3(9x + 4) > 35x - 4$

Explanation:

Let's solve each inequality step by step:

Problem 1: \( 3x - 1 > 3 + x \)

Step 1: Subtract \( x \) from both sides

\( 3x - x - 1 > 3 + x - x \)
\( 2x - 1 > 3 \)

Step 2: Add 1 to both sides

\( 2x - 1 + 1 > 3 + 1 \)
\( 2x > 4 \)

Step 3: Divide by 2

\( \frac{2x}{2} > \frac{4}{2} \)
\( x > 2 \)

Problem 2: \( \frac{1}{2}(x - 4) > x + 8 \)

Step 1: Distribute \( \frac{1}{2} \)

\( \frac{1}{2}x - 2 > x + 8 \)

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

\( \frac{1}{2}x - \frac{1}{2}x - 2 > x - \frac{1}{2}x + 8 \)
\( -2 > \frac{1}{2}x + 8 \)

Step 3: Subtract 8 from both sides

\( -2 - 8 > \frac{1}{2}x + 8 - 8 \)
\( -10 > \frac{1}{2}x \)

Step 4: Multiply by 2

\( -20 > x \) or \( x < -20 \)

Problem 3: \( 1 - 8x \geq 3x - 10 \)

Step 1: Add \( 8x \) to both sides

\( 1 - 8x + 8x \geq 3x + 8x - 10 \)
\( 1 \geq 11x - 10 \)

Step 2: Add 10 to both sides

\( 1 + 10 \geq 11x - 10 + 10 \)
\( 11 \geq 11x \)

Step 3: Divide by 11

\( \frac{11}{11} \geq \frac{11x}{11} \)
\( 1 \geq x \) or \( x \leq 1 \)

Problem 4: \( 2x + 6 < 5x - 3 \)

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

\( 2x - 2x + 6 < 5x - 2x - 3 \)
\( 6 < 3x - 3 \)

Step 2: Add 3 to both sides

\( 6 + 3 < 3x - 3 + 3 \)
\( 9 < 3x \)

Step 3: Divide by 3

\( 3 < x \) or \( x > 3 \)

Problem 5: \( -4(x - 8) \geq -2(2x - 16) \)

Step 1: Distribute both sides

\( -4x + 32 \geq -4x + 32 \)

Step 2: Add \( 4x \) to both sides

\( -4x + 4x + 32 \geq -4x + 4x + 32 \)
\( 32 \geq 32 \)

This is always true, so all real numbers \( x \) are solutions.

Problem 6: \( 3x - 2 \leq 2x + 1 \)

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

\( 3x - 2x - 2 \leq 2x - 2x + 1 \)
\( x - 2 \leq 1 \)

Step 2: Add 2 to both sides

\( x - 2 + 2 \leq 1 + 2 \)
\( x \leq 3 \)

Problem 7: \( -2(8 - 3x) > 6x + 2 \)

Step 1: Distribute \( -2 \)

\( -16 + 6x > 6x + 2 \)

Step 2: Subtract \( 6x \) from both sides

\( -16 + 6x - 6x > 6x - 6x + 2 \)
\( -16 > 2 \)

This is never true, so no solution.

Problem 8: \( -3(3x + 5) > -5(2x - 2) \)

Step 1: Distribute both sides

\( -9x - 15 > -10x + 10 \)

Step 2: Add \( 10x \) to both sides

\( -9x + 10x - 15 > -10x + 10x + 10 \)
\( x - 15 > 10 \)

Step 3: Add 15 to both sides

\( x - 15 + 15 > 10 + 15 \)
\( x > 25 \)

Problem 9: \( 8(2x - 1) > 11x + 17 \)

Step 1: Distribute \( 8 \)

\( 16x - 8 > 11x + 17 \)

Step 2: Subtract \( 11x \) from both sides

\( 16x - 11x - 8 > 11x - 11x + 17 \)
\( 5x - 8 > 17 \)

Step 3: Add 8 to both sides

\( 5x - 8 + 8 > 17 + 8 \)
\( 5x > 25 \)

Step 4: Divide by 5

\( \frac{5x}{5} > \frac{25}{5} \)
\( x > 5 \)

Problem 10: \( 3(9x + 4) > 35x - 4 \)

Step 1: Distribute \( 3 \)

\( 27x + 12 > 35x - 4 \)

Step 2: Subtract \( 27x \) from both sides

\( 27x - 27x + 12 > 35x - 27x - 4 \)
\( 12 > 8x - 4 \)

Step 3: Add 4 to both sides

\( 12 + 4 > 8x - 4 + 4 \)
\( 16 > 8x \)

Step 4: Divide by 8

\( \frac{16}{8} > \frac{8x}{8} \)
\( 2 > x \) or \( x < 2 \)

Final Answers:

1. \( \boldsymbol{x > 2} \)
2. \( \boldsymbol{x < -20} \)
3. \( \boldsymbol{x \leq 1} \)
4. \( \boldsymbol{x > 3} \)
5. \( \boldsymbol{\text{All real numbers}} \)
6. \( \boldsymbol{x \leq 3} \)
7. \( \boldsymbol{\text{No solution}} \)
8. \( \boldsymbol{x > 25} \)
9. \( \boldsymbol{x > 5} \)
10. \( \boldsymbol{x < 2} \)

Answer:

Let's solve each inequality step by step:

Problem 1: \( 3x - 1 > 3 + x \)

Step 1: Subtract \( x \) from both sides

\( 3x - x - 1 > 3 + x - x \)
\( 2x - 1 > 3 \)

Step 2: Add 1 to both sides

\( 2x - 1 + 1 > 3 + 1 \)
\( 2x > 4 \)

Step 3: Divide by 2

\( \frac{2x}{2} > \frac{4}{2} \)
\( x > 2 \)

Problem 2: \( \frac{1}{2}(x - 4) > x + 8 \)

Step 1: Distribute \( \frac{1}{2} \)

\( \frac{1}{2}x - 2 > x + 8 \)

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

\( \frac{1}{2}x - \frac{1}{2}x - 2 > x - \frac{1}{2}x + 8 \)
\( -2 > \frac{1}{2}x + 8 \)

Step 3: Subtract 8 from both sides

\( -2 - 8 > \frac{1}{2}x + 8 - 8 \)
\( -10 > \frac{1}{2}x \)

Step 4: Multiply by 2

\( -20 > x \) or \( x < -20 \)

Problem 3: \( 1 - 8x \geq 3x - 10 \)

Step 1: Add \( 8x \) to both sides

\( 1 - 8x + 8x \geq 3x + 8x - 10 \)
\( 1 \geq 11x - 10 \)

Step 2: Add 10 to both sides

\( 1 + 10 \geq 11x - 10 + 10 \)
\( 11 \geq 11x \)

Step 3: Divide by 11

\( \frac{11}{11} \geq \frac{11x}{11} \)
\( 1 \geq x \) or \( x \leq 1 \)

Problem 4: \( 2x + 6 < 5x - 3 \)

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

\( 2x - 2x + 6 < 5x - 2x - 3 \)
\( 6 < 3x - 3 \)

Step 2: Add 3 to both sides

\( 6 + 3 < 3x - 3 + 3 \)
\( 9 < 3x \)

Step 3: Divide by 3

\( 3 < x \) or \( x > 3 \)

Problem 5: \( -4(x - 8) \geq -2(2x - 16) \)

Step 1: Distribute both sides

\( -4x + 32 \geq -4x + 32 \)

Step 2: Add \( 4x \) to both sides

\( -4x + 4x + 32 \geq -4x + 4x + 32 \)
\( 32 \geq 32 \)

This is always true, so all real numbers \( x \) are solutions.

Problem 6: \( 3x - 2 \leq 2x + 1 \)

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

\( 3x - 2x - 2 \leq 2x - 2x + 1 \)
\( x - 2 \leq 1 \)

Step 2: Add 2 to both sides

\( x - 2 + 2 \leq 1 + 2 \)
\( x \leq 3 \)

Problem 7: \( -2(8 - 3x) > 6x + 2 \)

Step 1: Distribute \( -2 \)

\( -16 + 6x > 6x + 2 \)

Step 2: Subtract \( 6x \) from both sides

\( -16 + 6x - 6x > 6x - 6x + 2 \)
\( -16 > 2 \)

This is never true, so no solution.

Problem 8: \( -3(3x + 5) > -5(2x - 2) \)

Step 1: Distribute both sides

\( -9x - 15 > -10x + 10 \)

Step 2: Add \( 10x \) to both sides

\( -9x + 10x - 15 > -10x + 10x + 10 \)
\( x - 15 > 10 \)

Step 3: Add 15 to both sides

\( x - 15 + 15 > 10 + 15 \)
\( x > 25 \)

Problem 9: \( 8(2x - 1) > 11x + 17 \)

Step 1: Distribute \( 8 \)

\( 16x - 8 > 11x + 17 \)

Step 2: Subtract \( 11x \) from both sides

\( 16x - 11x - 8 > 11x - 11x + 17 \)
\( 5x - 8 > 17 \)

Step 3: Add 8 to both sides

\( 5x - 8 + 8 > 17 + 8 \)
\( 5x > 25 \)

Step 4: Divide by 5

\( \frac{5x}{5} > \frac{25}{5} \)
\( x > 5 \)

Problem 10: \( 3(9x + 4) > 35x - 4 \)

Step 1: Distribute \( 3 \)

\( 27x + 12 > 35x - 4 \)

Step 2: Subtract \( 27x \) from both sides

\( 27x - 27x + 12 > 35x - 27x - 4 \)
\( 12 > 8x - 4 \)

Step 3: Add 4 to both sides

\( 12 + 4 > 8x - 4 + 4 \)
\( 16 > 8x \)

Step 4: Divide by 8

\( \frac{16}{8} > \frac{8x}{8} \)
\( 2 > x \) or \( x < 2 \)

Final Answers:

1. \( \boldsymbol{x > 2} \)
2. \( \boldsymbol{x < -20} \)
3. \( \boldsymbol{x \leq 1} \)
4. \( \boldsymbol{x > 3} \)
5. \( \boldsymbol{\text{All real numbers}} \)
6. \( \boldsymbol{x \leq 3} \)
7. \( \boldsymbol{\text{No solution}} \)
8. \( \boldsymbol{x > 25} \)
9. \( \boldsymbol{x > 5} \)
10. \( \boldsymbol{x < 2} \)