Solve Systems Exactly

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Answer Key & Explanations

Solve Systems Exactly · Grade 9 · Worksheet 2

1. Solve the system: 3x + 2y = 7 and 5x - 4y = 2 Answer: x = 2, y = 0.5 Solution: Multiply the first equation by 2 to make the y-coefficients opposites: 2(3x + 2y) = 2(7) → 6x + 4y = 14 Add this to the second equation: (6x + 4y) + (5x - 4y) = 14 + 2 → 11x = 16 → x = 16/11 Substitute x = 16/11 into the first equation: 3(16/11) + 2y = 7 → 48/11 + 2y = 7 Solve for y: 2y = 7 -…
   Full step-by-step solution

   Step 1: Multiply the first equation by 2 to make the y-coefficients opposites: 2(3x + 2y) = 2(7) → 6x + 4y = 14 Step 2: Add this to the second equation: (6x + 4y) + (5x - 4y) = 14 + 2 → 11x = 16 → x = 16/11 Step 3: Substitute x = 16/11 into the first equation: 3(16/11) + 2y = 7 → 48/11 + 2y = 7 Step 4: Solve for y: 2y = 7 - 48/11 = 77/11 - 48/11 = 29/11 → y = 29/22 Step 5: The solution is x = 16/11, y = 29/22

2. Solve the system: y = x² - 9x + 20 and y = 2x - 10. Answer: x = 5, y = 0 and x = 6, y = 2 Solution: Set the equations equal: x² - 9x + 20 = 2x - 10. Bring all terms to one side: x² - 9x - 2x + 20 + 10 = 0 → x² - 11x + 30 = 0. Factor the quadratic: (x - 5)(x - 6) = 0.
   Full step-by-step solution

   Step 1: Set the equations equal: x² - 9x + 20 = 2x - 10. Step 2: Bring all terms to one side: x² - 9x - 2x + 20 + 10 = 0 → x² - 11x + 30 = 0. Step 3: Factor the quadratic: (x - 5)(x - 6) = 0. Step 4: Solve for x: x = 5 or x = 6. Step 5: Substitute x = 5 into y = 2x - 10: y = 2(5) - 10 = 10 - 10 = 0. Step 6: Substitute x = 6 into y = 2x - 10: y = 2(6) - 10 = 12 - 10 = 2. Step 7: Verify with the first equation: For x = 5, y = 25 - 45 + 20 = 0 ✓; for x = 6, y = 36 - 54 + 20 = 2 ✓. The solutions are (5, 0) and (6, 2).

3. Solve the system: 3x - 2y = 7 and 5x + 4y = 8 Answer: x = 2, y = -1/2 Solution: Step 1: Multiply the first equation by 2 to make the y-coefficients opposites: 2(3x - 2y) = 2(7) → 6x - 4y = 14 Step 2: Add this to the second equation: (6x - 4y) + (5x + 4y) = 14 + 8 → 11x = 22 Step 3: Solve for x: x = 22/11 = 2 Step 4: Substitute x = 2 into the first equation: 3(2) - 2y = 7 →…
   Full step-by-step solution

   Step 1: Multiply the first equation by 2 to make the y-coefficients opposites: 2(3x - 2y) = 2(7) → 6x - 4y = 14 Step 2: Add this to the second equation: (6x - 4y) + (5x + 4y) = 14 + 8 → 11x = 22 Step 3: Solve for x: x = 22/11 = 2 Step 4: Substitute x = 2 into the first equation: 3(2) - 2y = 7 → 6 - 2y = 7 Step 5: Solve for y: -2y = 1 → y = -1/2 Step 6: Verify in second equation: 5(2) + 4(-1/2) = 10 - 2 = 8 ✓ The solution is x = 2, y = -1/2.

4. 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 wants to create 500 mL of a mixture that contains exactly 30% acid. Write and solve a system of equations to determine how many milliliters of each solution should be used. Answer: x = 200, y = 300 Solution: When solving mixture problems, it's helpful to create two equations: one for the total quantity of the mixture and another for the total amount of the substance being mixed (in this case, acid).
   Full step-by-step solution

   When solving mixture problems, it's helpful to create two equations: one for the total quantity of the mixture and another for the total amount of the substance being mixed (in this case, acid). The second equation comes from multiplying each solution's volume by its concentration percentage (expressed as a decimal). This approach can be applied to various mixture scenarios, such as mixing different grades of fuel or combining materials with different purity levels.

5. 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 ml of a mixture that contains exactly 30% acid. Write and solve a system of equations to determine how many milliliters of each solution should be used. Answer: 200 ml of Solution A and 300 ml of Solution B Solution: Let x be the volume of Solution A (15% acid) and y be the volume of Solution B (40% acid).
   Full step-by-step solution

   Step 1: Let x be the volume of Solution A (15% acid) and y be the volume of Solution B (40% acid). Step 2: Set up the volume equation: x + y = 500 Step 3: Set up the acid content equation: 0.15x + 0.40y = 0.30(500) Step 4: Simplify the acid equation: 0.15x + 0.40y = 150 Step 5: Solve the system using substitution. From the first equation: y = 500 - x Step 6: Substitute into the acid equation: 0.15x + 0.40(500 - x) = 150 Step 7: Expand: 0.15x + 200 - 0.40x = 150 Step 8: Combine like terms: -0.25x + 200 = 150 Step 9: Subtract 200 from both sides: -0.25x = -50 Step 10: Divide both sides by -0.25: x = 200 Step 11: Substitute back to find y: y = 500 - 200 = 300 Step 12: The lab should use 200 ml of Solution A and 300 ml of Solution B.

6. Solve the system: 3x + 2y = 16 and 5x - 3y = -5 Answer: x = 2, y = 5 Solution: Multiply the first equation by 3 and the second by 2 to eliminate y (3x + 2y = 16) × 3 → 9x + 6y = 48 (5x - 3y = -5) × 2 → 10x - 6y = -10 9x + 6y = 48 10x - 6y = -10 19x = 38 x = 38 ÷ 19 x = 2 Substitute x = 2 into the first equation 3(2) + 2y = 16 6 + 2y = 16 2y = 10 y = 5 5(2) - 3(5) = 10 - 15…
   Full step-by-step solution

   Step 1: Multiply the first equation by 3 and the second by 2 to eliminate y (3x + 2y = 16) × 3 → 9x + 6y = 48 (5x - 3y = -5) × 2 → 10x - 6y = -10 Step 2: Add the two equations to eliminate y 9x + 6y = 48 10x - 6y = -10 19x = 38 Step 3: Solve for x x = 38 ÷ 19 x = 2 Step 4: Substitute x = 2 into the first equation 3(2) + 2y = 16 6 + 2y = 16 2y = 10 y = 5 Step 5: Verify the solution in the second equation 5(2) - 3(5) = 10 - 15 = -5 ✓ The solution is x = 2, y = 5.

7. A chemistry lab needs to mix two solutions of hydrochloric acid. Solution A is 25% acid and Solution B is 40% acid. They want to obtain 60 liters of a mixture that is 30% acid. Write and solve a system of equations to determine how many liters of each solution should be used. Answer: 40 liters of Solution A and 20 liters of Solution B Solution: Let x = liters of Solution A (25% acid) Let y = liters of Solution B (40% acid) Total volume equation: x + y = 60 Acid content equation: 0.25x + 0.40y = 0.30(60) Simplify acid equation: 0.25x + 0.40y = 18 Solve the system using substitution: y = 60 - x Substitute into acid equation: 0.25x +…
   Full step-by-step solution

   Step 1: Let x = liters of Solution A (25% acid) Step 2: Let y = liters of Solution B (40% acid) Step 3: Total volume equation: x + y = 60 Step 4: Acid content equation: 0.25x + 0.40y = 0.30(60) Step 5: Simplify acid equation: 0.25x + 0.40y = 18 Step 6: Solve the system using substitution: y = 60 - x Step 7: Substitute into acid equation: 0.25x + 0.40(60 - x) = 18 Step 8: Expand: 0.25x + 24 - 0.40x = 18 Step 9: Combine like terms: -0.15x + 24 = 18 Step 10: Subtract 24: -0.15x = -6 Step 11: Divide by -0.15: x = 40 Step 12: Find y: y = 60 - 40 = 20 Step 13: Check: 0.25(40) + 0.40(20) = 10 + 8 = 18, which equals 0.30(60) The answer is 40 liters of Solution A and 20 liters of Solution B.

8. Liam is designing a rectangular garden with a perimeter of 40 meters. He wants the length to be 4 meters more than twice the width. Write and solve a system of equations to determine the exact dimensions of Liam's garden. Answer: width = 8 meters, length = 12 meters Solution: When solving geometry problems with given relationships between dimensions, we often use the perimeter formula along with the relationship equation. For rectangles, perimeter equals twice the sum of length and width.
   Full step-by-step solution

   When solving geometry problems with given relationships between dimensions, we often use the perimeter formula along with the relationship equation. For rectangles, perimeter equals twice the sum of length and width. The relationship between dimensions gives us a second equation that allows us to substitute and solve for one variable first, then find the other.