QUESTION IMAGE
Question
- $3x - 1 > 3 + x$
$+1 +1$
$3x > 4 + x$
$-x \\ \\ \\ \\ \\ -x$
$\frac{2x}{2} > \frac{4}{2}, \\ x > 2$
- $\frac{1}{2}(x - 4) > x + 8$
$\frac{1}{2}x - 2 > x + 8$
$-x \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ -x$
$-0.5x - 2 > 8$
$\\ \\ \\ \\ \\ \\ \\ +2 \\ \\ \\ +2$
$\\ \\ \\ \\ \\ \\ -0.5x > 10$
$\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ x < -20$
- $1 - 8x \geq 3x - 10$
$-3x -3x$
$1 - 11x \geq -10$
$-1 \\ \\ \\ \\ \\ \\ \\ \\ \\ -1$
$\frac{-11x}{-11} \geq \frac{-11}{-11}$
$x \leq 1$
- $2x + 6 < 5x - 3$
$-5x \\ \\ \\ \\ -5x$
$-3x + 6 < -3$
$\\ \\ \\ \\ \\ \\ -6 \\ \\ \\ -6$
$\frac{-3x}{-3} < \frac{-9}{-3}$
$\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ x > 3$
- $-4(x - 8) \geq -2(2x - 16)$
- $3x - 2 \leq 2x + 1$
Let's solve each inequality one by one:
1. Solve \( 3x - 1 > 3 + x \)
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 both sides by 2
\( \frac{2x}{2} > \frac{4}{2} \)
\( x > 2 \)
2. Solve \( \frac{1}{2}(x - 4) > x + 8 \)
Step 1: Multiply both sides by 2 to eliminate the fraction
\( 2\times\frac{1}{2}(x - 4) > 2\times(x + 8) \)
\( x - 4 > 2x + 16 \)
Step 2: Subtract \( x \) from both sides
\( x - x - 4 > 2x - x + 16 \)
\( -4 > x + 16 \)
Step 3: Subtract 16 from both sides
\( -4 - 16 > x + 16 - 16 \)
\( -20 > x \) or \( x < -20 \)
3. Solve \( 1 - 8x \geq 3x - 10 \)
Step 1: Subtract \( 3x \) from both sides
\( 1 - 8x - 3x \geq 3x - 3x - 10 \)
\( 1 - 11x \geq -10 \)
Step 2: Subtract 1 from both sides
\( 1 - 1 - 11x \geq -10 - 1 \)
\( -11x \geq -11 \)
Step 3: Divide both sides by -11 (remember to reverse the inequality sign)
\( \frac{-11x}{-11} \leq \frac{-11}{-11} \)
\( x \leq 1 \)
4. Solve \( 2x + 6 < 5x - 3 \)
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 both sides by 3
\( \frac{9}{3} < \frac{3x}{3} \)
\( 3 < x \) or \( x > 3 \)
5. Solve \( -4(x - 8) \geq -2(2x - 16) \)
Step 1: Expand 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 the solution is all real numbers.
6. Solve \( 3x - 2 \leq 2x + 1 \)
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
\( x - 2 + 2 \leq 1 + 2 \)
\( x \leq 3 \)
Final Answers:
- \( \boldsymbol{x > 2} \)
- \( \boldsymbol{x < -20} \)
- \( \boldsymbol{x \leq 1} \)
- \( \boldsymbol{x > 3} \)
- \( \boldsymbol{\text{All real numbers}} \)
- \( \boldsymbol{x \leq 3} \)
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Let's solve each inequality one by one:
1. Solve \( 3x - 1 > 3 + x \)
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 both sides by 2
\( \frac{2x}{2} > \frac{4}{2} \)
\( x > 2 \)
2. Solve \( \frac{1}{2}(x - 4) > x + 8 \)
Step 1: Multiply both sides by 2 to eliminate the fraction
\( 2\times\frac{1}{2}(x - 4) > 2\times(x + 8) \)
\( x - 4 > 2x + 16 \)
Step 2: Subtract \( x \) from both sides
\( x - x - 4 > 2x - x + 16 \)
\( -4 > x + 16 \)
Step 3: Subtract 16 from both sides
\( -4 - 16 > x + 16 - 16 \)
\( -20 > x \) or \( x < -20 \)
3. Solve \( 1 - 8x \geq 3x - 10 \)
Step 1: Subtract \( 3x \) from both sides
\( 1 - 8x - 3x \geq 3x - 3x - 10 \)
\( 1 - 11x \geq -10 \)
Step 2: Subtract 1 from both sides
\( 1 - 1 - 11x \geq -10 - 1 \)
\( -11x \geq -11 \)
Step 3: Divide both sides by -11 (remember to reverse the inequality sign)
\( \frac{-11x}{-11} \leq \frac{-11}{-11} \)
\( x \leq 1 \)
4. Solve \( 2x + 6 < 5x - 3 \)
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 both sides by 3
\( \frac{9}{3} < \frac{3x}{3} \)
\( 3 < x \) or \( x > 3 \)
5. Solve \( -4(x - 8) \geq -2(2x - 16) \)
Step 1: Expand 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 the solution is all real numbers.
6. Solve \( 3x - 2 \leq 2x + 1 \)
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
\( x - 2 + 2 \leq 1 + 2 \)
\( x \leq 3 \)
Final Answers:
- \( \boldsymbol{x > 2} \)
- \( \boldsymbol{x < -20} \)
- \( \boldsymbol{x \leq 1} \)
- \( \boldsymbol{x > 3} \)
- \( \boldsymbol{\text{All real numbers}} \)
- \( \boldsymbol{x \leq 3} \)