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(g) Solve for \( x \) and represent your answer on a number line: \( \begin{array}{ll}\text { (1) } 2 x+1 \geq 13 & \text { (2) } 2-2 x \leq 3 \\ \text { (3) } 3(x+7)>6 & \text { (4) } 10-2 x<22+2 x \\ \text { (5) } \frac{1-x}{2} \leq 3-x & \text { (6) } \frac{-5 x}{3}-1 \geq x+7\end{array} \)

Ask by Savage Ramirez. in South Africa
Mar 05,2025

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Answer

The solution is \( x \geq 6 \).

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The Deep Dive

To tackle each inequality, we will isolate \( x \) and solve them step-by-step: 1. **For \( 2x + 1 \geq 13 \):** - Subtract 1 from both sides: \( 2x \geq 12 \) - Divide by 2: \( x \geq 6 \) 2. **For \( 2 - 2x \leq 3 \):** - Subtract 2 from both sides: \( -2x \leq 1 \) - Divide by -2 (reverse the inequality): \( x \geq -0.5 \) 3. **For \( 3(x + 7) > 6 \):** - Distribute: \( 3x + 21 > 6 \) - Subtract 21 from both sides: \( 3x > -15 \) - Divide by 3: \( x > -5 \) 4. **For \( 10 - 2x < 22 + 2x \):** - Add \( 2x \) to both sides: \( 10 < 22 + 4x \) - Subtract 22 from both sides: \( -12 < 4x \) - Divide by 4: \( x > -3 \) 5. **For \( \frac{1 - x}{2} \leq 3 - x \):** - Multiply by 2: \( 1 - x \leq 6 - 2x \) - Add \( x \) to both sides: \( 1 \leq 6 - x \) - Adding \( x \): \( x \leq 5 \) 6. **For \( \frac{-5x}{3} - 1 \geq x + 7 \):** - Multiply through by 3: \( -5x - 3 \geq 3x + 21 \) - Add \( 5x \) to both sides: \( -3 \geq 8x + 21 \) - Subtract 21: \( -24 \geq 8x \) - Divide by 8: \( x \leq -3 \) Now we combine all the results we got for each inequality: 1. \( x \geq 6 \) 2. \( x \geq -0.5 \) 3. \( x > -5 \) 4. \( x > -3 \) 5. \( x \leq 5 \) 6. \( x \leq -3 \) To summarize, the relationships can be simplified as follows: - From inequalities \( 1 \) and \( 5 \), the solution for \( x \) must fulfill both conditions, so we find that \( x \) can take any value starting from 6 onward: \( x \ge 6 \). - The stricter condition among \( 6 \) and \( 4 \) is \( x \le -3 \) but reasonably attainable only in a specific intersection. To represent on a number line: - Shade the area to the right from \( 6 \) (inclusive) and to the left from \( -3\) (inclusive). You end up with two distinct shaded areas on your number line at points \( -3 \) and \( 6 \).

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