Solve Systems Exactly

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Solve Systems Exactly · Grade 9 · Worksheet 3

1. On a coordinate plane, Aroha draws a parabola with equation y = x^2 - 7 and a line with equation y = 2x + 1. Visualize these two graphs intersecting. What are the exact coordinates of their intersection points? Answer: (4, 9) and (-2, -3) Solution: Set the equations equal to each other since at intersection points, y is the same for both: x^2 - 7 = 2x + 1 Rearrange to standard form: x^2 - 7 - 2x - 1 = 0 → x^2 - 2x - 8 = 0 Factor the quadratic: (x - 4)(x + 2) = 0 Solve for x: x - 4 = 0 gives x = 4, and x + 2 = 0 gives x = -2 Find the…
   Full step-by-step solution

   Step 1: Set the equations equal to each other since at intersection points, y is the same for both: x^2 - 7 = 2x + 1 Step 2: Rearrange to standard form: x^2 - 7 - 2x - 1 = 0 → x^2 - 2x - 8 = 0 Step 3: Factor the quadratic: (x - 4)(x + 2) = 0 Step 4: Solve for x: x - 4 = 0 gives x = 4, and x + 2 = 0 gives x = -2 Step 5: Find the corresponding y-coordinates using y = 2x + 1: When x = 4: y = 2(4) + 1 = 8 + 1 = 9 → (4, 9) When x = -2: y = 2(-2) + 1 = -4 + 1 = -3 → (-2, -3) Step 6: Verify with the parabola equation: For (4, 9): 4^2 - 7 = 16 - 7 = 9 ✓ For (-2, -3): (-2)^2 - 7 = 4 - 7 = -3 ✓ The exact coordinates of the intersection points are (4, 9) and (-2, -3).

2. Solve the system: y = x² - 6x + 11 and y = 2x - 1. Answer: x = 4, y = 7 and x = 3, y = 5 Solution: Step 1: Set the equations equal: x² - 6x + 11 = 2x - 1 Step 2: Rearrange to standard form: x² - 6x - 2x + 11 + 1 = 0 → x² - 8x + 12 = 0 Step 3: Factor the quadratic: (x - 6)(x - 2) = 0 Step 4: Solve for x: x - 6 = 0 → x = 6; x - 2 = 0 → x = 2 Step 5: Substitute x = 6 into y = 2x - 1: y = 2(6) -…
   Full step-by-step solution

   Step 1: Set the equations equal: x² - 6x + 11 = 2x - 1 Step 2: Rearrange to standard form: x² - 6x - 2x + 11 + 1 = 0 → x² - 8x + 12 = 0 Step 3: Factor the quadratic: (x - 6)(x - 2) = 0 Step 4: Solve for x: x - 6 = 0 → x = 6; x - 2 = 0 → x = 2 Step 5: Substitute x = 6 into y = 2x - 1: y = 2(6) - 1 = 12 - 1 = 11 Step 6: Substitute x = 2 into y = 2x - 1: y = 2(2) - 1 = 4 - 1 = 3 Step 7: Verify with the quadratic: For x = 6: 6² - 6(6) + 11 = 36 - 36 + 11 = 11 ✓; For x = 2: 2² - 6(2) + 11 = 4 - 12 + 11 = 3 ✓ The solutions are x = 6, y = 11 and x = 2, y = 3.

3. Solve the system: 3x² - 2x + 5 = y and 2x² + 4x - 1 = y Answer: x = -3, x = 2 Solution: When solving systems of quadratic equations, set the expressions equal since they both equal y. This creates a new equation that can be solved by moving all terms to one side and factoring or using the quadratic formula.
   Full step-by-step solution

   When solving systems of quadratic equations, set the expressions equal since they both equal y. This creates a new equation that can be solved by moving all terms to one side and factoring or using the quadratic formula. Quadratic equations often have two solutions, so be sure to find all possible x-values.

4. Solve the system: 3x² - 2x + 5 = 2x² + 3x - 1 and 4x - y = 7 Answer: x = 2, y = 1 Solution: Simplify the first equation: 3x² - 2x + 5 = 2x² + 3x - 1 Subtract 2x² from both sides: x² - 2x + 5 = 3x - 1 Subtract 3x from both sides: x² - 5x + 5 = -1 Add 1 to both sides: x² - 5x + 6 = 0 Factor the quadratic: (x - 2)(x - 3) = 0 So x = 2 or x = 3 Use the second equation 4x - y = 7 For x = 2:…
   Full step-by-step solution

   Step 1: Simplify the first equation: 3x² - 2x + 5 = 2x² + 3x - 1 Subtract 2x² from both sides: x² - 2x + 5 = 3x - 1 Subtract 3x from both sides: x² - 5x + 5 = -1 Add 1 to both sides: x² - 5x + 6 = 0 Step 2: Factor the quadratic: (x - 2)(x - 3) = 0 So x = 2 or x = 3 Step 3: Use the second equation 4x - y = 7 For x = 2: 4(2) - y = 7 → 8 - y = 7 → y = 1 For x = 3: 4(3) - y = 7 → 12 - y = 7 → y = 5 Step 4: Verify both solutions in the original equations: For (2,1): 3(4) - 4 + 5 = 13 and 2(4) + 6 - 1 = 13 ✓ For (3,5): 3(9) - 6 + 5 = 26 and 2(9) + 9 - 1 = 26 ✓ Both pairs satisfy the system, so the solutions are x = 2, y = 1 and x = 3, y = 5.

5. Solve the system: y = x² - 5x + 7 and y = 3x - 5 Answer: x = 3, y = 4 and x = 4, y = 7 Solution: Step 1: Set the equations equal: x² - 5x + 7 = 3x - 5 Step 2: Rearrange to standard form: x² - 5x - 3x + 7 + 5 = 0 → x² - 8x + 12 = 0 Step 3: Factor the quadratic: (x - 2)(x - 6) = 0 Step 4: Solve for x: x - 2 = 0 → x = 2; x - 6 = 0 → x = 6 Step 5: Substitute x = 2 into y = 3x - 5: y = 3(2) - 5…
   Full step-by-step solution

   Step 1: Set the equations equal: x² - 5x + 7 = 3x - 5 Step 2: Rearrange to standard form: x² - 5x - 3x + 7 + 5 = 0 → x² - 8x + 12 = 0 Step 3: Factor the quadratic: (x - 2)(x - 6) = 0 Step 4: Solve for x: x - 2 = 0 → x = 2; x - 6 = 0 → x = 6 Step 5: Substitute x = 2 into y = 3x - 5: y = 3(2) - 5 = 6 - 5 = 1 Step 6: Substitute x = 6 into y = 3x - 5: y = 3(6) - 5 = 18 - 5 = 13 Step 7: Verify with the first equation: For x = 2: 2² - 5(2) + 7 = 4 - 10 + 7 = 1 ✓; For x = 6: 6² - 5(6) + 7 = 36 - 30 + 7 = 13 ✓ The solutions are (x = 2, y = 1) and (x = 6, y = 13).

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. 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.

8. Solve the system: 3x² - 2x + y = 7 and 2x + y = 5 Answer: x = 2, y = 1 and x = -1/3, y = 17/3 Solution: From the second equation 2x + y = 5, solve for y: y = 5 - 2x Substitute y = 5 - 2x into the first equation: 3x² - 2x + (5 - 2x) = 7 Simplify: 3x² - 2x + 5 - 2x = 7 → 3x² - 4x + 5 = 7 Subtract 7 from both sides: 3x² - 4x - 2 = 0 Use the quadratic formula: x = [4 ± √(16 + 24)]/6 = [4 ± √40]/6 = [4…
   Full step-by-step solution

   Step 1: From the second equation 2x + y = 5, solve for y: y = 5 - 2x Step 2: Substitute y = 5 - 2x into the first equation: 3x² - 2x + (5 - 2x) = 7 Step 3: Simplify: 3x² - 2x + 5 - 2x = 7 → 3x² - 4x + 5 = 7 Step 4: Subtract 7 from both sides: 3x² - 4x - 2 = 0 Step 5: Use the quadratic formula: x = [4 ± √(16 + 24)]/6 = [4 ± √40]/6 = [4 ± 2√10]/6 = [2 ± √10]/3 Step 6: For x = (2 + √10)/3: y = 5 - 2(2 + √10)/3 = 5 - (4 + 2√10)/3 = (15 - 4 - 2√10)/3 = (11 - 2√10)/3 Step 7: For x = (2 - √10)/3: y = 5 - 2(2 - √10)/3 = 5 - (4 - 2√10)/3 = (15 - 4 + 2√10)/3 = (11 + 2√10)/3 The solutions are x = (2 + √10)/3, y = (11 - 2√10)/3 and x = (2 - √10)/3, y = (11 + 2√10)/3

9. 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.