Solve Systems Exactly

lessonbunny.com

Answer Key & Explanations

Solve Systems Exactly · Grade 9 · Worksheet 1

1. A right triangle is drawn on a coordinate plane with vertices at (0,0), (6,0), and (0,8). A circle is circumscribed around this triangle, passing through all three vertices. Visualize the triangle inside the circle with the circle's circumference touching each vertex. What is the area of the circumscribed circle? Answer: 25π Solution: Identify the right triangle vertices: (0,0), (6,0), and (0,8) The hypotenuse is the line segment from (6,0) to (0,8) Calculate the length of the hypotenuse using the distance formula: sqrt((6-0)^2 + (0-8)^2) = sqrt(36 + 64) = sqrt(100) = 10 For a right triangle, the hypotenuse is the diameter of…
   Full step-by-step solution

   Step 1: Identify the right triangle vertices: (0,0), (6,0), and (0,8) Step 2: The hypotenuse is the line segment from (6,0) to (0,8) Step 3: Calculate the length of the hypotenuse using the distance formula: sqrt((6-0)^2 + (0-8)^2) = sqrt(36 + 64) = sqrt(100) = 10 Step 4: For a right triangle, the hypotenuse is the diameter of the circumscribed circle, so diameter = 10 Step 5: Radius = diameter/2 = 10/2 = 5 Step 6: Area of circle = π × radius^2 = π × 5^2 = 25π Step 7: The area is 25π square units

2. Matiu is designing a rectangular garden that is surrounded by a uniform gravel path. The garden itself has a length that is 4 meters more than its width. The total area of the garden and the path together is 140 square meters. The path is 2 meters wide on all sides. Find the exact dimensions (length and width) of the garden. Answer: Width = 6 meters, Length = 10 meters Solution: Let w be the width of the garden in meters. Then the length of the garden is w + 4 meters. The path is 2 meters wide on all sides, so the total width of garden plus path is w + 2 + 2 = w + 4 meters.
   Full step-by-step solution

   Step 1: Let w be the width of the garden in meters. Then the length of the garden is w + 4 meters. Step 2: The path is 2 meters wide on all sides, so the total width of garden plus path is w + 2 + 2 = w + 4 meters. The total length is (w + 4) + 2 + 2 = w + 8 meters. Step 3: The total area (garden + path) is given as 140 square meters, so (w + 4)(w + 8) = 140. Step 4: Expand the left side: w^2 + 8w + 4w + 32 = 140, which simplifies to w^2 + 12w + 32 = 140. Step 5: Subtract 140 from both sides: w^2 + 12w + 32 - 140 = 0, so w^2 + 12w - 108 = 0. Step 6: Solve the quadratic equation. Factor: (w + 18)(w - 6) = 0. Step 7: So w = -18 or w = 6. Since width cannot be negative, w = 6 meters. Step 8: Then the length is w + 4 = 6 + 4 = 10 meters. Step 9: Check: Total width = 6 + 4 = 10 meters, total length = 10 + 8 = 18 meters. Area = 10 * 18 = 140 square meters. Correct. The garden is 6 meters wide and 10 meters long.

3. On a coordinate plane, Liam draws a parabola that opens upward with vertex at (3, 1) and a line that passes through the points (1, 5) and (5, 5). The parabola and the line intersect at two points. Visualize this graph. What are the exact coordinates of the two intersection points? Answer: (1, 5) and (5, 5) Solution: The parabola has vertex (3, 1). Its equation in vertex form is y = a(x - 3)^2 + 1. Since it opens upward, a > 0.
   Full step-by-step solution

   Step 1: The parabola has vertex (3, 1). Its equation in vertex form is y = a(x - 3)^2 + 1. Since it opens upward, a > 0. We need to find a. The line passes through (1, 5) and (5, 5), so its slope is (5-5)/(5-1) = 0. The line is horizontal at y = 5. Step 2: The parabola must pass through the same points where it meets the line. Substitute (1, 5) into the parabola equation: 5 = a(1 - 3)^2 + 1 => 5 = a(4) + 1 => 4 = 4a => a = 1. So the parabola is y = (x - 3)^2 + 1. Step 3: Set the equations equal: (x - 3)^2 + 1 = 5 => (x - 3)^2 = 4 => x - 3 = 2 or x - 3 = -2 => x = 5 or x = 1. Step 4: The y-coordinate for both is 5. The intersection points are (1, 5) and (5, 5).

4. A chemistry lab needs to mix two solutions of hydrochloric acid. Solution A is 25% acid and Solution B is 50% acid. They want to create 8 liters of a 40% acid solution. Write a system of equations to model this situation and solve it exactly to determine how many liters of each solution should be used. Answer: 3.2 liters of Solution A and 4.8 liters of Solution B Solution: Let x = liters of Solution A (25% acid) and y = liters of Solution B (50% acid) Create the volume equation: x + y = 8 Create the acid content equation: 0.25x + 0.50y = 0.40(8) Simplify the acid equation: 0.25x + 0.50y = 3.2 Solve the system using substitution: y = 8 - x Substitute into the acid…
   Full step-by-step solution

   Step 1: Let x = liters of Solution A (25% acid) and y = liters of Solution B (50% acid) Step 2: Create the volume equation: x + y = 8 Step 3: Create the acid content equation: 0.25x + 0.50y = 0.40(8) Step 4: Simplify the acid equation: 0.25x + 0.50y = 3.2 Step 5: Solve the system using substitution: y = 8 - x Step 6: Substitute into the acid equation: 0.25x + 0.50(8 - x) = 3.2 Step 7: Simplify: 0.25x + 4 - 0.50x = 3.2 Step 8: Combine like terms: -0.25x + 4 = 3.2 Step 9: Subtract 4 from both sides: -0.25x = -0.8 Step 10: Divide by -0.25: x = 3.2 Step 11: Find y: y = 8 - 3.2 = 4.8 Step 12: The solution is 3.2 liters of Solution A and 4.8 liters of Solution B

5. x² + y² = 85, y = 2x - 5. Solve system. Answer: x = 6, y = 7 and x = -4, y = -13 Solution: Substitute y = 2x - 5 into x² + y² = 85: x² + (2x - 5)² = 85. Expand (2x - 5)² = 4x² - 20x + 25. Combine: x² + 4x² - 20x + 25 = 85 → 5x² - 20x + 25 = 85.
   Full step-by-step solution

   Step 1: Substitute y = 2x - 5 into x² + y² = 85: x² + (2x - 5)² = 85. Step 2: Expand (2x - 5)² = 4x² - 20x + 25. Step 3: Combine: x² + 4x² - 20x + 25 = 85 → 5x² - 20x + 25 = 85. Step 4: Subtract 85: 5x² - 20x - 60 = 0. Step 5: Divide by 5: x² - 4x - 12 = 0. Step 6: Factor: (x - 6)(x + 4) = 0 → x = 6 or x = -4. Step 7: For x = 6, y = 2(6) - 5 = 12 - 5 = 7. Step 8: For x = -4, y = 2(-4) - 5 = -8 - 5 = -13. Step 9: Verify: (6)² + (7)² = 36 + 49 = 85 ✓; (-4)² + (-13)² = 16 + 169 = 85 ✓. The solutions are (6, 7) and (-4, -13).

6. A chemistry lab needs to mix two solutions with different concentrations of acid. Solution A contains 15% acid and Solution B contains 40% acid. The lab wants to create 200 milliliters of a mixture that contains exactly 30% acid. Let x represent the amount of Solution A in milliliters and y represent the amount of Solution B in milliliters. Write and solve a system of equations to determine exactly how many milliliters of each solution should be used. Answer: x = 80, y = 120 Solution: When solving mixture problems, we typically set up two equations: one representing the total quantity (like total volume) and another representing the total amount of the substance being mixed (like pure acid).
   Full step-by-step solution

   When solving mixture problems, we typically set up two equations: one representing the total quantity (like total volume) and another representing the total amount of the substance being mixed (like pure acid). The concentration percentage tells us what fraction of each solution is the pure substance. For example, if you had 100 ml of a 20% solution, it would contain 20 ml of pure substance.

7. A chemical company is testing two different purification methods for a new compound. Method A removes contaminants at a rate modeled by the equation y = 2x + 5, where y is the percentage of contaminants removed after x minutes. Method B follows the equation y = -x² + 12x. After how many minutes will both methods remove the same percentage of contaminants, and what is that percentage? Answer: 5 minutes and 15% Solution: This creates a system where we solve for the variable that makes both equations true simultaneously.
   Full step-by-step solution

   When comparing two different mathematical models that represent real-world processes, we can find when they produce equal results by setting the equations equal to each other. This creates a system where we solve for the variable that makes both equations true simultaneously. In practical applications, we must consider which solutions are physically meaningful within the context of the problem.