Create Inequalities

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Answer Key & Explanations

Create Inequalities · Grade 9 · Worksheet 3

1. Aisha is designing a rectangular banner for a school event. The banner must have a perimeter of at most 36 feet. She wants the length to be at least 3 feet more than twice the width. If w represents the width of the banner in feet, write and solve an inequality to determine all possible values for w that satisfy Aisha's design requirements. Answer: 0 < w ≤ 5 Solution: Let w = width (in feet). The length is at least 3 more than twice the width, so length ≥ 2w + 3. The perimeter of a rectangle is 2(length + width).
   Full step-by-step solution

   Step 1: Let w = width (in feet). The length is at least 3 more than twice the width, so length ≥ 2w + 3. Step 2: The perimeter of a rectangle is 2(length + width). The perimeter must be at most 36 feet, so 2(length + width) ≤ 36. Step 3: Substitute the minimum length expression: 2((2w + 3) + w) ≤ 36. Step 4: Simplify inside the parentheses: 2(3w + 3) ≤ 36. Step 5: Distribute the 2: 6w + 6 ≤ 36. Step 6: Subtract 6 from both sides: 6w ≤ 30. Step 7: Divide both sides by 6: w ≤ 5. Step 8: Since width must be positive: w > 0. Step 9: Combine both conditions: 0 < w ≤ 5. The possible width values are greater than 0 feet and at most 5 feet.

2. 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

3. Lily is planning a school fundraiser and needs to raise at least $408. Each raffle ticket sells for $6. Write an inequality to represent how many tickets Lily must sell to meet the goal. Answer: 68 Solution: Let x = number of tickets sold. Each ticket sells for $6, so revenue = 6x. We need revenue ≥ 408, so 6x ≥ 408.
   Full step-by-step solution

   Let x = number of tickets sold. Each ticket sells for $6, so revenue = 6x. We need revenue ≥ 408, so 6x ≥ 408. Divide both sides by 6: x ≥ 408 ÷ 6 = 68. Since Lily can only sell whole tickets, Lily must sell at least 68 tickets. Inequality: x ≥ 68.

4. A right triangle is inscribed in a semicircle with the hypotenuse as the diameter. The semicircle has a radius of 6 units. A rectangle is drawn inside the triangle such that one side lies along the hypotenuse and the opposite vertices touch the two legs of the triangle. If the width of the rectangle (along the hypotenuse) is represented by x, write an inequality that must be true for x based on the geometric constraints of the triangle. Answer: 0 < x < 12 Solution: Since the triangle is inscribed in a semicircle with the hypotenuse as the diameter, it is a right triangle with the right angle at the vertex opposite the hypotenuse. The hypotenuse is the diameter of the semicircle, so its length is 2 × 6 = 12 units.
   Full step-by-step solution

   Step 1: Since the triangle is inscribed in a semicircle with the hypotenuse as the diameter, it is a right triangle with the right angle at the vertex opposite the hypotenuse. Step 2: The hypotenuse is the diameter of the semicircle, so its length is 2 × 6 = 12 units. Step 3: Let the legs of the triangle be a and b. By the Pythagorean theorem, a² + b² = 12² = 144. Step 4: The rectangle has width x along the hypotenuse and touches both legs. The height of the rectangle will be determined by similar triangles. Step 5: Consider the two smaller right triangles formed at the ends of the rectangle. They are similar to the original triangle. Step 6: The sum of the bases of these two smaller triangles equals the hypotenuse minus the rectangle's width: (base1 + base2) = 12 - x. Step 7: Since the rectangle touches both legs, its height h must satisfy h > 0 and be less than the maximum possible height of the triangle. Step 8: For the rectangle to exist inside the triangle, we must have x > 0 and 12 - x > 0, which gives 0 < x < 12. Step 9: Therefore, the inequality that must be true is 0 < x < 12.

5. Mason is saving money to buy a new laptop that costs $1,200. He already has $320 saved and plans to save $55 each week from his part-time job. What is the minimum number of weeks he needs to save to afford the laptop? Write an inequality to represent this situation and solve for the number of weeks. Answer: w ≥ 16 Solution: Let w represent the number of weeks Mason saves. Each week he saves $55, so after w weeks he saves 55w dollars. He already has $320, so total savings = 320 + 55w.
   Full step-by-step solution

   Step 1: Let w represent the number of weeks Mason saves. Step 2: Each week he saves $55, so after w weeks he saves 55w dollars. Step 3: He already has $320, so total savings = 320 + 55w. Step 4: He needs at least $1,200, so the inequality is: 320 + 55w ≥ 1200. Step 5: Subtract 320 from both sides: 55w ≥ 880. Step 6: Divide both sides by 55: w ≥ 16. Step 7: Since w represents weeks, the minimum number of whole weeks is 16. Final answer: w ≥ 16.

6. A rectangular garden has a length that is 5 meters more than twice its width. The area of the garden must be at least 42 square meters. If the width is represented by w meters, write an inequality in terms of w that represents this situation, then solve for the possible values of w. Answer: w ≥ 3.5 Solution: Define the variables and relationship between length and width. Let \( w \) = width of the garden in meters. The length \( l \) is 5 meters more than twice the width: l = 2w + 5 Write the area condition.
   Full step-by-step solution

   Let's go step by step. --- **Step 1: Define the variables and relationship between length and width.** Let \( w \) = width of the garden in meters. The length \( l \) is 5 meters more than twice the width: \[ l = 2w + 5 \] --- **Step 2: Write the area condition.** Area = length × width = \( l \times w \) Substitute \( l \): \[ \text{Area} = (2w + 5) \times w \] The area must be at least 42 square meters: \[ (2w + 5) \times w \geq 42 \] --- **Step 3: Expand and rearrange into a standard inequality.** \[ 2w^2 + 5w \geq 42 \] Subtract 42 from both sides: \[ 2w^2 + 5w - 42 \geq 0 \] --- **Step 4: Solve the quadratic equation \( 2w^2 + 5w - 42 = 0 \).** Use the quadratic formula: \[ w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here \( a = 2 \), \( b = 5 \), \( c = -42 \). \[ w = \frac{-5 \pm \sqrt{25 - 4(2)(-42)}}{2(2)} \] \[ w = \frac{-5 \pm \sqrt{25 + 336}}{4} \] \[ w = \frac{-5 \pm \sqrt{361}}{4} \] \[ w = \frac{-5 \pm 19}{4} \] --- **Step 5: Find the two roots.** First root: \[ w = \frac{-5 + 19}{4} = \frac{14}{4} = 3.5 \] Second root: \[ w = \frac{-5 - 19}{4} = \frac{-24}{4} = -6 \] --- **Step 6: Interpret the inequality \( 2w^2 + 5w - 42 \geq 0 \).** The quadratic opens upward (coefficient of \( w^2 \) is positive), so it is ≥ 0 outside the roots. But \( w \) is a width, so \( w > 0 \). Thus the relevant interval is \( w \geq 3.5 \). --- **Step 7: Final answer.** \[ w \geq 3.5 \]

7. A right triangle is inscribed in a circle such that the hypotenuse is the diameter of the circle. The legs of the triangle are represented by the expressions (2x - 1) cm and (x + 3) cm, and the hypotenuse is 13 cm. Write an inequality that represents all possible values of x for which the triangle's perimeter is greater than 30 cm. Answer: x > 2 Solution: When working with geometric inequalities involving triangles, remember that all side lengths must be positive. The perimeter inequality comes from summing the expressions for the sides.
   Full step-by-step solution

   When working with geometric inequalities involving triangles, remember that all side lengths must be positive. The perimeter inequality comes from summing the expressions for the sides. Additionally, the triangle inequality theorem states that the sum of any two sides must be greater than the third side, which provides important constraints on the variable.