Create Inequalities

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Create Inequalities · Grade 9 · Worksheet 2

1. Liam is designing a rectangular garden with a perimeter of 40 meters. He wants the length of the garden to be at least 5 meters more than twice the width. Write an inequality in terms of the width w that represents this situation, then determine the possible range of widths for Liam's garden. Answer: w ≤ 5 Solution: - Perimeter = 40 meters - Let \( w \) = width (in meters) - Let \( l \) = length (in meters) P = 2l + 2w Given \( P = 40 \): 2l + 2w = 40 Divide through by 2: l + w = 20 l = 20 - w The length is at least 5 meters more than twice the width: l \geq 2w + 5 Substitute \( l = 20 - w \) into the…
   Full step-by-step solution

   Let's go step-by-step. --- **Step 1: Understand the problem** We have a rectangular garden with: - Perimeter = 40 meters - Let \( w \) = width (in meters) - Let \( l \) = length (in meters) Perimeter formula for a rectangle: \[ P = 2l + 2w \] Given \( P = 40 \): \[ 2l + 2w = 40 \] Divide through by 2: \[ l + w = 20 \] So: \[ l = 20 - w \] --- **Step 2: Translate the condition** The length is at least 5 meters more than twice the width: \[ l \geq 2w + 5 \] --- **Step 3: Substitute \( l = 20 - w \) into the inequality** \[ 20 - w \geq 2w + 5 \] --- **Step 4: Solve for \( w \)** \[ 20 - w \geq 2w + 5 \] \[ 20 - 5 \geq 2w + w \] \[ 15 \geq 3w \] \[ 3w \leq 15 \] \[ w \leq 5 \] --- **Step 5: Consider physical constraints** Width must be positive: \( w > 0 \). Also, from \( l = 20 - w \), if \( w \leq 5 \), then \( l \geq 15 \), which is fine. So the possible range of widths is: \[ 0 < w \leq 5 \] (Since width must be positive for a garden.) --- **Final answer:** \[ w \leq 5 \] (assuming \( w > 0 \) is understood for a real garden)

2. 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.

3. 2x² - 8x + 6 < 0 Answer: 1 < x < 3 Solution: 2x² - 8x + 6 < 0 All coefficients are divisible by 2: 2(x² - 4x + 3) < 0 We need two numbers that multiply to 3 and add to -4: Those numbers are -1 and -3.
   Full step-by-step solution

   Let's solve the inequality step by step. We have: 2x² - 8x + 6 < 0 --- **Step 1: Factor out the common factor** All coefficients are divisible by 2: 2(x² - 4x + 3) < 0 --- **Step 2: Factor the quadratic inside** We need two numbers that multiply to 3 and add to -4: Those numbers are -1 and -3. So: x² - 4x + 3 = (x - 1)(x - 3) Thus: 2(x - 1)(x - 3) < 0 --- **Step 3: Divide both sides by 2** Since 2 > 0, dividing by 2 does not change the inequality direction: (x - 1)(x - 3) < 0 --- **Step 4: Find the critical points** Set (x - 1)(x - 3) = 0 x = 1 or x = 3 These are the points where the expression equals zero. --- **Step 5: Determine intervals** The critical points split the number line into three intervals: (-∞, 1), (1, 3), (3, ∞) --- **Step 6: Test each interval** We want (x - 1)(x - 3) < 0 (negative). - For x in (-∞, 1): pick x = 0 (0 - 1)(0 - 3) = (-1)(-3) = 3 > 0 → positive → not part of solution. - For x in (1, 3): pick x = 2 (2 - 1)(2 - 3) = (1)(-1) = -1 < 0 → negative → satisfies inequality. - For x in (3, ∞): pick x = 4 (4 - 1)(4 - 3) = (3)(1) = 3 > 0 → positive → not part of solution. --- **Step 7: Check endpoints** At x = 1 or x = 3, expression = 0, but we need < 0 (strict inequality), so endpoints are not included. --- **Step 8: Write the solution** The expression is negative only on (1, 3). So: 1 < x < 3 --- **Final answer:** 1 < x < 3

4. 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.

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 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 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 inequality 2w² + 3w - 54 ≥ 0 is satisfied when w ≤ -6 or w ≥ 4.5. Step 7: Considering w > 0, the solution is w ≥ 4.5. The answer is w ≥ 4.5.

6. Mason is organizing a charity fun run. Each participant needs to run at least 50 kilometers. If Mason has already run 27 kilometers, write an inequality representing the additional kilometers, x, they must still run to meet the goal. Answer: 23 Solution: The goal is to run at least 50 kilometers, so the total distance must be greater than or equal to 50. Let x be the additional kilometers needed. Write the inequality: 27 + x ≥ 50.
   Full step-by-step solution

   Step 1: The goal is to run at least 50 kilometers, so the total distance must be greater than or equal to 50. Step 2: Let x be the additional kilometers needed. Step 3: Write the inequality: 27 + x ≥ 50. Step 4: Solve for x: x ≥ 50 - 27. Step 5: x ≥ 23 kilometers. The answer is x ≥ 23.

7. Liam is designing a rectangular garden that will be inscribed inside a circular flower bed with a diameter of 10 meters. The length of the garden is represented by x meters, and the width is 3 meters less than the length. The garden must have an area of at least 28 square meters. Write an inequality in terms of x that represents this area constraint, and then determine the possible values of x that satisfy both the geometric constraint (the garden must fit inside the circle) and the area constraint. Answer: 3.5 ≤ x ≤ 8 Solution: Let the length be x meters and the width be (x - 3) meters. The diagonal of the rectangle equals the diameter of the circle, which is 10 meters. By the Pythagorean theorem: x^2 + (x - 3)^2 = 10^2.
   Full step-by-step solution

   Step 1: Let the length be x meters and the width be (x - 3) meters. The diagonal of the rectangle equals the diameter of the circle, which is 10 meters. By the Pythagorean theorem: x^2 + (x - 3)^2 = 10^2. Step 2: Expand: x^2 + x^2 - 6x + 9 = 100 → 2x^2 - 6x + 9 = 100 → 2x^2 - 6x - 91 = 0. Solve using the quadratic formula: x = [6 ± sqrt(36 + 728)] / 4 = [6 ± sqrt(764)] / 4 = [6 ± 2*sqrt(191)] / 4 = (3 ± sqrt(191)) / 2. sqrt(191) is about 13.82, so the positive root is about (3 + 13.82)/2 = 8.41. Since length must be positive and less than the diagonal, x must be less than 10. For the rectangle to fit inside the circle, the diagonal must be exactly 10, so x is uniquely determined by the equation above. However, the problem asks for inequality: the rectangle must fit inside the circle, meaning the diagonal must be less than or equal to 10. So: x^2 + (x - 3)^2 ≤ 100. Step 3: Simplify: 2x^2 - 6x + 9 ≤ 100 → 2x^2 - 6x - 91 ≤ 0. The quadratic equals zero at x = (3 ± sqrt(191))/2. The positive root is about 8.41, and the negative root is about -5.41. Since x > 0, the inequality holds for 0 < x ≤ 8.41. Step 4: Area inequality: area = x(x - 3) ≥ 28 → x^2 - 3x - 28 ≥ 0. Factor: (x - 7)(x + 4) ≥ 0. The zeros are x = 7 and x = -4. Since x > 0, the inequality holds for x ≥ 7. Step 5: Combine the constraints: x must satisfy x ≥ 7 and x ≤ 8.41. Also, width x - 3 > 0 → x > 3, which is already satisfied. Therefore, the possible values are 7 ≤ x ≤ 8.41. Since x represents a length in a real-world context, and the diagonal constraint gives an exact maximum, we round to the nearest tenth: the inequality is 7 ≤ x ≤ 8.4. However, using exact values: x must be between 7 and (3 + sqrt(191))/2. The problem asks for the inequality in terms of x, so the final answer is: x ≥ 7 and x ≤ (3 + sqrt(191))/2. In simplest radical form, this is 7 ≤ x ≤ (3 + sqrt(191))/2. Approximating, the answer is 7 ≤ x ≤ 8.4, but the exact answer is 7 ≤ x ≤ (3 + sqrt(191))/2. Since the problem asks for possible values, the answer is: 7 ≤ x ≤ 8.4 (approximately).