Solve Systems Exactly

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Answer Key & Explanations

Solve Systems Exactly · Grade 9 · Worksheet 1

1. Solve the system: 3x - 2y = 7 and 2x + 5y = -4 Answer: x = 1, y = -2 Solution: Multiply the first equation by 5 and the second by 2 to eliminate y: 5(3x - 2y) = 5(7) → 15x - 10y = 35 2(2x + 5y) = 2(-4) → 4x + 10y = -8 (15x - 10y) + (4x + 10y) = 35 + (-8) 15x + 4x - 10y + 10y = 27 19x = 27 x = 27/19 x = 1 Substitute x = 1 into the first equation: 3(1) - 2y = 7 3 - 2y = 7…
   Full step-by-step solution

   Step 1: Multiply the first equation by 5 and the second by 2 to eliminate y: 5(3x - 2y) = 5(7) → 15x - 10y = 35 2(2x + 5y) = 2(-4) → 4x + 10y = -8 Step 2: Add the two equations: (15x - 10y) + (4x + 10y) = 35 + (-8) 15x + 4x - 10y + 10y = 27 19x = 27 Step 3: Solve for x: x = 27/19 x = 1 Step 4: Substitute x = 1 into the first equation: 3(1) - 2y = 7 3 - 2y = 7 -2y = 7 - 3 -2y = 4 y = -2 Step 5: Verify with the second equation: 2(1) + 5(-2) = 2 - 10 = -8 ✓ The solution is x = 1, y = -2.

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. A chemistry lab needs to mix two solutions with different concentrations of hydrochloric acid. Solution A contains 15% acid and Solution B contains 40% acid. The lab technician wants to create 500 milliliters of a solution that contains exactly 28% acid. Write and solve a system of equations to determine how many milliliters of each solution should be mixed. Answer: 240 ml of Solution A and 260 ml of Solution B Solution: Step 1: Let x = milliliters of Solution A (15% acid) Step 2: Let y = milliliters of Solution B (40% acid) Step 3: Create the volume equation: x + y = 500 Step 4: Create the acid equation: 0.15x + 0.40y = 0.28(500) Step 5: Simplify the acid equation: 0.15x + 0.40y = 140 Step 6: Solve the system…
   Full step-by-step solution

   Step 1: Let x = milliliters of Solution A (15% acid) Step 2: Let y = milliliters of Solution B (40% acid) Step 3: Create the volume equation: x + y = 500 Step 4: Create the acid equation: 0.15x + 0.40y = 0.28(500) Step 5: Simplify the acid equation: 0.15x + 0.40y = 140 Step 6: Solve the system using substitution: y = 500 - x Step 7: Substitute into acid equation: 0.15x + 0.40(500 - x) = 140 Step 8: Simplify: 0.15x + 200 - 0.40x = 140 Step 9: Combine like terms: -0.25x + 200 = 140 Step 10: Subtract 200 from both sides: -0.25x = -60 Step 11: Divide by -0.25: x = 240 Step 12: Find y: y = 500 - 240 = 260 Step 13: Verify: 240 + 260 = 500 ml total, and 0.15(240) + 0.40(260) = 36 + 104 = 140 ml acid, which is 28% of 500 ml Therefore, the lab needs 240 ml of Solution A and 260 ml of Solution B.

4. 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 500 milliliters of a mixture that contains exactly 30% acid. Write a system of equations to model this situation and solve it exactly to determine how many milliliters of each solution should be used. Answer: A = 200, B = 300 Solution: Define variables. Let A be the volume of Solution A (15% acid) and B be the volume of Solution B (40% acid). Write the volume equation.
   Full step-by-step solution

   Step 1: Define variables. Let A be the volume of Solution A (15% acid) and B be the volume of Solution B (40% acid). Step 2: Write the volume equation. The total volume is 500 ml: A + B = 500 Step 3: Write the acid equation. The acid from Solution A is 0.15A, from Solution B is 0.40B, and the total acid in the mixture is 0.30 × 500 = 150 ml: 0.15A + 0.40B = 150 Step 4: Solve the system. From A + B = 500, we get A = 500 - B Step 5: Substitute into the acid equation: 0.15(500 - B) + 0.40B = 150 Step 6: Simplify: 75 - 0.15B + 0.40B = 150 Step 7: Combine like terms: 75 + 0.25B = 150 Step 8: Subtract 75 from both sides: 0.25B = 75 Step 9: Divide both sides by 0.25: B = 300 Step 10: Substitute back to find A: A = 500 - 300 = 200 Step 11: Check: 0.15(200) + 0.40(300) = 30 + 120 = 150, which equals 0.30 × 500 = 150 The answer is A = 200, B = 300.

5. 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

6. A rectangular garden has a length that is 3 meters more than twice its width. The area of the garden is 65 square meters. Find the width of the garden in meters. Answer: 5 Solution: Let the width of the garden be \( w \) meters. The length is 3 meters more than twice the width, so: length \( l = 2w + 3 \).
   Full step-by-step solution

   Let's go step-by-step. --- **Step 1: Define variables** Let the width of the garden be \( w \) meters. The length is 3 meters more than twice the width, so: length \( l = 2w + 3 \). --- **Step 2: Write the area equation** Area of rectangle = length × width Given area = 65 m², so: \( l \times w = 65 \) Substitute \( l = 2w + 3 \): \( (2w + 3) \times w = 65 \). --- **Step 3: Expand and rearrange** \( 2w^2 + 3w = 65 \) Subtract 65 from both sides: \( 2w^2 + 3w - 65 = 0 \). --- **Step 4: Solve the quadratic equation** Use the quadratic formula: \( w = \frac{ -b \pm \sqrt{b^2 - 4ac} }{ 2a } \) Here \( a = 2 \), \( b = 3 \), \( c = -65 \). First, discriminant: \( D = b^2 - 4ac = 3^2 - 4(2)(-65) \) \( D = 9 + 520 = 529 \). --- **Step 5: Take square root of discriminant** \( \sqrt{529} = 23 \). --- **Step 6: Apply quadratic formula** \( w = \frac{ -3 \pm 23 }{ 2 \times 2 } \) \( w = \frac{ -3 \pm 23 }{ 4 } \). Two possible solutions: \( w = \frac{ -3 + 23 }{ 4 } = \frac{20}{4} = 5 \) \( w = \frac{ -3 - 23 }{ 4 } = \frac{-26}{4} = -6.5 \). --- **Step 7: Choose the valid solution** Width cannot be negative, so \( w = -6.5 \) is rejected. Thus \( w = 5 \) meters. --- **Step 8: Check** Width \( w = 5 \) m Length \( l = 2(5) + 3 = 13 \) m Area = \( 5 \times 13 = 65 \) m² — correct. --- **Final answer:** 5

7. 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. What is the area of the circumscribed circle? (Use π = 3.14) Answer: 78.5 Solution: Identify that this is a right triangle with legs of length 6 and 8 units For a right triangle, the hypotenuse is the diameter of the circumscribed circle Calculate the hypotenuse using Pythagorean theorem: sqrt(6^2 + 8^2) = sqrt(36 + 64) = sqrt(100) = 10 The diameter of the circle is 10, so the…
   Full step-by-step solution

   Step 1: Identify that this is a right triangle with legs of length 6 and 8 units Step 2: For a right triangle, the hypotenuse is the diameter of the circumscribed circle Step 3: Calculate the hypotenuse using Pythagorean theorem: sqrt(6^2 + 8^2) = sqrt(36 + 64) = sqrt(100) = 10 Step 4: The diameter of the circle is 10, so the radius is 10 ÷ 2 = 5 Step 5: Calculate the area of the circle using A = πr^2 = 3.14 × 5^2 = 3.14 × 25 = 78.5 The answer is 78.5.