Create Inequalities

lessonbunny.com

Answer Key & Explanations

Create Inequalities · Grade 9 · Worksheet 2

1. Matiu is saving to buy a new bicycle that costs $455. He already has $95 and plans to save $45 each week. What is the minimum number of weeks he needs to save to afford the bicycle? Write an inequality and solve. Answer: w ≥ 8 Solution: Let w represent the number of weeks Matiu saves. He already has $95 and saves $45 per week, so total saved = 95 + 45w. He needs at least $455, so the inequality is 95 + 45w ≥ 455.
   Full step-by-step solution

   Step 1: Let w represent the number of weeks Matiu saves. Step 2: He already has $95 and saves $45 per week, so total saved = 95 + 45w. Step 3: He needs at least $455, so the inequality is 95 + 45w ≥ 455. Step 4: Subtract 95 from both sides: 45w ≥ 360. Step 5: Divide both sides by 45: w ≥ 8. Step 6: Since w must be a whole number, the minimum number of weeks is 8. Final answer: w ≥ 8.

2. Aisha is designing a rectangular vegetable patch with a fixed perimeter of 80 meters. She wants the length to be at least 10 meters more than three times the width. If w represents the width in meters, which inequality represents all possible widths that satisfy Aisha's design requirements? Answer: C. 2w + 2(3w + 10) ≥ 80 Solution: Let w be the width in meters. The length is at least 10 meters more than three times the width, so length ≥ 3w + 10. The perimeter of a rectangle is given by P = 2(length) + 2(width).
   Full step-by-step solution

   Step 1: Let w be the width in meters. Step 2: The length is at least 10 meters more than three times the width, so length ≥ 3w + 10. Step 3: The perimeter of a rectangle is given by P = 2(length) + 2(width). Step 4: The perimeter is fixed at 80 meters, so 2(length) + 2(width) = 80. Step 5: Since the length must be at least (3w + 10), the perimeter must be at least 2(3w + 10) + 2w. Step 6: Write the inequality: 2(3w + 10) + 2w ≥ 80. Step 7: Simplify the left side: 6w + 20 + 2w = 8w + 20. Step 8: The inequality becomes 8w + 20 ≥ 80. Step 9: The correct representation before simplification is 2w + 2(3w + 10) ≥ 80, which corresponds to choice C.

3. 3x² - 12x + 9 ≤ 0 Answer: 1 ≤ x ≤ 3 Solution: 3x² - 12x + 9 ≤ 0 All terms are divisible by 3: 3(x² - 4x + 3) ≤ 0 x² - 4x + 3 factors into (x - 1)(x - 3). 3(x - 1)(x - 3) ≤ 0 x - 1 = 0 → x = 1 x - 3 = 0 → x = 3 These are the points where the product is zero.
   Full step-by-step solution

   Let's solve the inequality step by step. We have: 3x² - 12x + 9 ≤ 0 --- **Step 1: Factor out the greatest common factor** All terms are divisible by 3: 3(x² - 4x + 3) ≤ 0 --- **Step 2: Factor the quadratic inside** x² - 4x + 3 factors into (x - 1)(x - 3). So we have: 3(x - 1)(x - 3) ≤ 0 --- **Step 3: Identify critical points** The expression equals zero when: x - 1 = 0 → x = 1 x - 3 = 0 → x = 3 These are the points where the product is zero. --- **Step 4: Determine the sign of the expression in each interval** The factor 3 is positive, so it doesn't affect the sign. We check the sign of (x - 1)(x - 3): - For x < 1: pick x = 0 (0 - 1) = -1, (0 - 3) = -3 (-1)(-3) = 3 > 0 → positive - For 1 < x < 3: pick x = 2 (2 - 1) = 1 > 0, (2 - 3) = -1 < 0 (+)(-) = - → negative - For x > 3: pick x = 4 (4 - 1) = 3 > 0, (4 - 3) = 1 > 0 (+)(+) = + → positive --- **Step 5: Apply the inequality sign** We want 3(x - 1)(x - 3) ≤ 0. Since 3 > 0, this means (x - 1)(x - 3) ≤ 0. From the sign chart: Negative region is between 1 and 3, and zero at x = 1 and x = 3. So solution is: 1 ≤ x ≤ 3 --- **Final answer:** 1 ≤ x ≤ 3

4. Aisha is designing a rectangular mural for her school's art competition. The mural's length must be at least 3 meters more than twice its width. If the total area of the mural must be at least 54 square meters, and the width is represented by w meters, write and solve an inequality to determine the possible width values for Aisha's mural. Answer: w ≥ 4.5 Solution: Let w represent the width in meters. The length is at least 3 more than twice the width, so length ≥ 2w + 3. The area must be at least 54 square meters, so (length)(width) ≥ 54.
   Full step-by-step solution

   Step 1: Let w represent the width in meters. Step 2: The length is at least 3 more than twice the width, so length ≥ 2w + 3. Step 3: The area must be at least 54 square meters, so (length)(width) ≥ 54. Step 4: Substitute the expression for length: (2w + 3)(w) ≥ 54. Step 5: Expand the left side: 2w² + 3w ≥ 54. Step 6: Subtract 54 from both sides: 2w² + 3w - 54 ≥ 0. Step 7: Solve the quadratic equation 2w² + 3w - 54 = 0. Step 8: Use the quadratic formula: w = [-3 ± √(9 + 432)] / 4 = [-3 ± √441] / 4 = [-3 ± 21] / 4. Step 9: Calculate the two solutions: w = (-3 + 21)/4 = 18/4 = 4.5, and w = (-3 - 21)/4 = -24/4 = -6. Step 10: Since width must be positive, discard w = -6. Step 11: Test intervals: For w > 4.5, the inequality holds true. For 0 < w < 4.5, the inequality is false. Step 12: Therefore, the solution is w ≥ 4.5 meters.

5. Aisha is designing a rectangular banner for a school event. The banner's length must be at least 3 meters more than twice its width. If the area of the banner must be greater than 54 square meters, and the width is represented by w meters, write and solve an inequality to determine all possible widths for Aisha's banner. Answer: w > 4.5 Solution: Express the length in terms of width. Since length is at least 3 more than twice the width, length = 2w + 3. Write the area inequality.
   Full step-by-step solution

   Step 1: Express the length in terms of width. Since length is at least 3 more than twice the width, length = 2w + 3. Step 2: Write the area inequality. Area = length × width > 54, so (2w + 3) × w > 54. Step 3: Expand and rearrange: 2w² + 3w > 54 → 2w² + 3w - 54 > 0. Step 4: Solve the quadratic equation 2w² + 3w - 54 = 0 using the quadratic formula: w = [-3 ± √(9 + 432)] / 4 = [-3 ± √441] / 4 = [-3 ± 21] / 4. Step 5: Calculate the roots: w = (-3 + 21)/4 = 18/4 = 4.5, and w = (-3 - 21)/4 = -24/4 = -6. Step 6: Since width must be positive, discard w = -6. The parabola opens upward, so the inequality 2w² + 3w - 54 > 0 is satisfied when w < -6 or w > 4.5. Step 7: Considering only positive widths, the solution is w > 4.5.

6. Liam is designing a rectangular garden with a perimeter of 60 meters. He wants the length of the garden to be at least 5 meters more than twice the width. Write an inequality that represents all possible widths (in meters) that satisfy Liam's design requirements. Answer: 0 < w ≤ 25/3 or w ≤ 8.33 Solution: w = width of the garden (in meters) L = length of the garden (in meters) Perimeter of a rectangle = 2 × (length + width) Given: Perimeter = 60 meters 2(L + w) = 60 L + w = 30 L = 30 − w The problem says: length is at least 5 meters more than twice the width.
   Full step-by-step solution

   Let's go step by step. --- **Step 1: Define variables** Let w = width of the garden (in meters) L = length of the garden (in meters) --- **Step 2: Write the perimeter equation** Perimeter of a rectangle = 2 × (length + width) Given: Perimeter = 60 meters So: 2(L + w) = 60 L + w = 30 L = 30 − w --- **Step 3: Translate the condition on length** The problem says: length is at least 5 meters more than twice the width. "At least" means ≥. So: L ≥ 2w + 5 --- **Step 4: Substitute L from Step 2 into the inequality** 30 − w ≥ 2w + 5 --- **Step 5: Solve for w** 30 − w ≥ 2w + 5 30 − 5 ≥ 2w + w 25 ≥ 3w w ≤ 25/3 --- **Step 6: Consider the domain of w** Width must be positive: w > 0. Also, length L = 30 − w must be positive (since it's a rectangle), but that gives w < 30 automatically if w ≤ 25/3 and w > 0. So the possible widths are: 0 < w ≤ 25/3 --- **Step 7: Interpret the result** 25/3 = 8.333... (or 8.33 when rounded to two decimals). So the inequality is: 0 < w ≤ 25/3 --- **Final answer:** 0 < w ≤ 25/3 or w ≤ 8.33

7. Aisha is designing a rectangular mural for her school's art exhibition. The mural's length must be at least 3 meters more than twice its width. If the area of the mural needs to be at least 54 square meters, and the width is represented by w meters, write and solve an inequality to determine the possible width values for Aisha's mural. Answer: w ≥ 4.5 Solution: Express the length in terms of width. Since length is at least 3 more than twice the width: length = 2w + 3 Write the area inequality: (2w + 3) × w ≥ 54 Expand the left side: 2w² + 3w ≥ 54 Subtract 54 from both sides: 2w² + 3w - 54 ≥ 0 Solve the quadratic equation 2w² + 3w - 54 = 0 Use the…
   Full step-by-step solution

   Step 1: Express the length in terms of width. Since length is at least 3 more than twice the width: length = 2w + 3 Step 2: Write the area inequality: (2w + 3) × w ≥ 54 Step 3: Expand the left side: 2w² + 3w ≥ 54 Step 4: Subtract 54 from both sides: 2w² + 3w - 54 ≥ 0 Step 5: Solve the quadratic equation 2w² + 3w - 54 = 0 Step 6: Use the quadratic formula: w = [-3 ± √(9 + 432)] / 4 = [-3 ± √441] / 4 = [-3 ± 21] / 4 Step 7: Calculate the two solutions: w = (-3 + 21)/4 = 18/4 = 4.5, and w = (-3 - 21)/4 = -24/4 = -6 Step 8: Since width must be positive, we discard w = -6 Step 9: Test intervals: The inequality 2w² + 3w - 54 ≥ 0 is satisfied when w ≤ -6 or w ≥ 4.5 Step 10: Since width must be positive, the solution is w ≥ 4.5 The answer is w ≥ 4.5.