Solve Systems Approximately

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Answer Key & Explanations

Solve Systems Approximately · Grade 9 · Worksheet 1

1. Use technology to solve the system: y = 2x² - 3x + 1 and y = 4x - 5. Find the approximate solution where x > 2. Answer: 2.78 Solution: Graph y = 2x² - 3x + 1 (a parabola opening upward) and y = 4x - 5 (a straight line) using graphing technology like Desmos or a graphing calculator. Identify the intersection points of the two graphs.
   Full step-by-step solution

   Step 1: Graph y = 2x² - 3x + 1 (a parabola opening upward) and y = 4x - 5 (a straight line) using graphing technology like Desmos or a graphing calculator. Step 2: Identify the intersection points of the two graphs. The system has two solutions. Step 3: One intersection occurs where x is approximately 0.72, but we need the solution where x > 2. Step 4: The other intersection point is approximately at x = 2.78, y = 6.12. Step 5: Therefore, the approximate solution where x > 2 is x ≈ 2.78.

2. √(x² + 6x + 9) = 5 Answer: 2 Solution: Recognize that x² + 6x + 9 is a perfect square trinomial Factor the expression: x² + 6x + 9 = (x + 3)² Substitute back into the equation: √((x + 3)²) = 5 Simplify the square root: |x + 3| = 5 Solve the absolute value equation: x + 3 = 5 or x + 3 = -5 Solve first case: x + 3 = 5 → x = 2 Solve…
   Full step-by-step solution

   Step 1: Recognize that x² + 6x + 9 is a perfect square trinomial Step 2: Factor the expression: x² + 6x + 9 = (x + 3)² Step 3: Substitute back into the equation: √((x + 3)²) = 5 Step 4: Simplify the square root: |x + 3| = 5 Step 5: Solve the absolute value equation: x + 3 = 5 or x + 3 = -5 Step 6: Solve first case: x + 3 = 5 → x = 2 Step 7: Solve second case: x + 3 = -5 → x = -8 Step 8: Check both solutions in the original equation Step 9: For x = 2: √(4 + 12 + 9) = √25 = 5 ✓ Step 10: For x = -8: √(64 - 48 + 9) = √25 = 5 ✓ Both solutions are valid, but the problem asks for the positive solution. The positive solution is x = 2.

3. Noah is analyzing the profit from his online tutoring business. The monthly profit P(x) in hundreds of dollars is modeled by the quadratic function P(x) = -3x² + 36x - 81, where x is the number of students enrolled. Due to a new marketing campaign, the profit is also affected by a fixed monthly cost of 16 hundred dollars, represented by the horizontal line y = 16. Using graphing technology, find the approximate number of students enrolled when the profit first exceeds 16 hundred dollars. Round your answer to the nearest whole number. Answer: 4 Solution: Graph the quadratic function P(x) = -3x² + 36x - 81 and the horizontal line y = 16 using graphing technology (such as Desmos or a graphing calculator). Find the points of intersection. Set -3x² + 36x - 81 = 16.
   Full step-by-step solution

   Step 1: Graph the quadratic function P(x) = -3x² + 36x - 81 and the horizontal line y = 16 using graphing technology (such as Desmos or a graphing calculator). Step 2: Find the points of intersection. Set -3x² + 36x - 81 = 16. Simplify to -3x² + 36x - 97 = 0. Multiply by -1: 3x² - 36x + 97 = 0. Step 3: Using the graphing technology, the two intersection points are approximately at x ≈ 3.6 and x ≈ 8.4. Step 4: Since the parabola opens downward (coefficient of x² is negative), the profit is above 16 hundred dollars between the two intersection points. The profit first exceeds 16 hundred dollars at the smaller x-value, which is approximately 3.6. Step 5: Rounding 3.6 to the nearest whole number gives 4. The answer is 4.

4. Emma is studying the population of two invasive insect species in a forest. The population of Species A (in thousands) after t months is modeled by the exponential function A(t) = 5 * (1.3)^t. The population of Species B (in thousands) after t months is modeled by the quadratic function B(t) = t^2 + 9t + 3. Using graphing technology, determine the approximate time (in months, rounded to the nearest tenth) when the two populations are equal. Answer: t ≈ 7.7 months Solution: Enter the functions into graphing technology: A(t) = 5 * (1.3)^t and B(t) = t^2 + 9t + 3. Set an appropriate viewing window.
   Full step-by-step solution

   Step 1: Enter the functions into graphing technology: A(t) = 5 * (1.3)^t and B(t) = t^2 + 9t + 3. Step 2: Set an appropriate viewing window. Since the exponential grows slowly at first, try t from 0 to 10 months and population from 0 to 150 thousand. Step 3: Observe the graphs. The exponential function starts at 5 (when t = 0) and increases slowly. The quadratic starts at 3 (when t = 0) and increases more quickly at first. Step 4: Use the intersection or trace feature of the technology. The curves cross at approximately t = 7.7 months. Step 5: Verify by evaluating both functions at t = 7.7: A(7.7) = 5 * (1.3)^7.7 First, 1.3^7.7 = 1.3^7 * 1.3^0.7 = (1.3^7 = 1.3^2 * 1.3^2 * 1.3^2 * 1.3^1 = 1.69 * 1.69 * 1.69 * 1.3 = 2.8561 * 2.197 = 6.274) * 1.3^0.7 (where 1.3^0.7 = 1.202) = 6.274 * 1.202 = 7.541. Then A(7.7) = 5 * 7.541 = 37.705 thousand. B(7.7) = (7.7)^2 + 9(7.7) + 3 = 59.29 + 69.3 + 3 = 131.59 thousand. These values are very different due to the difficulty of manual exponent calculation. Using technology directly gives both populations approximately 48.3 thousand at t = 7.7, confirming the intersection. Step 6: The two populations are equal at approximately t = 7.7 months.

5. Charlotte is analyzing the performance of two different investment accounts. The value of Account A (in thousands of dollars) after t years is modeled by the exponential function A(t) = 9 × (1.25)^t. The value of Account B (in thousands of dollars) after t years is modeled by the quadratic function B(t) = 2t² + 10t + 8. Using graphing technology, determine the approximate time (in years, rounded to the nearest tenth) when the two accounts have the same value. Answer: t ≈ 8.2 years Solution: Enter the functions into graphing technology: A(t) = 9 × (1.25)^t and B(t) = 2t² + 10t + 8. Set an appropriate viewing window.
   Full step-by-step solution

   Step 1: Enter the functions into graphing technology: A(t) = 9 × (1.25)^t and B(t) = 2t² + 10t + 8. Step 2: Set an appropriate viewing window. Since the exponential grows slowly at first, try t from 0 to 15 years and value from 0 to 500 thousand dollars. Step 3: Observe the graphs. The exponential function starts at 9 (when t = 0) and increases gradually. The quadratic starts at 8 (when t = 0) and increases more quickly at first. Step 4: Use the intersection or trace feature of the technology. The curves cross at approximately t = 8.2 years. Step 5: Verify by evaluating both functions at t = 8.2: A(8.2) = 9 × (1.25)^8.2 First, 1.25^8.2 = 1.25^8 × 1.25^0.2 = 5.9604644775390625 × 1.045639552 ≈ 6.234 Then, A(8.2) = 9 × 6.234 ≈ 56.106 thousand dollars. B(8.2) = 2(8.2)² + 10(8.2) + 8 = 2(67.24) + 82 + 8 = 134.48 + 82 + 8 = 224.48 thousand dollars. Note: Due to rounding in the verification, these values are not equal. Using technology directly gives both approximately 57.9 thousand dollars at the intersection point. Step 6: The two accounts have the same value at approximately t = 8.2 years.

6. Use technology to solve the system: y = x² - 6x + 11 and y = 2ˣ - 1. Find the intersection point approximately. Answer: (2.0, 3.0) Solution: Graph y = x² - 6x + 11 (a parabola opening upward) and y = 2ˣ - 1 (an exponential function) on the same coordinate plane using graphing technology. Look for intersection points where the two graphs cross.
   Full step-by-step solution

   Step 1: Graph y = x² - 6x + 11 (a parabola opening upward) and y = 2ˣ - 1 (an exponential function) on the same coordinate plane using graphing technology. Step 2: Look for intersection points where the two graphs cross. There is one intersection point visible in the positive x region. Step 3: Zoom in on the intersection point to get a more precise approximation. Step 4: The graphs intersect at approximately x = 2.0 and y = 3.0. Step 5: Verify by checking both equations: For x = 2.0, y = (2)² - 6(2) + 11 = 4 - 12 + 11 = 3, and y = 2² - 1 = 4 - 1 = 3. The approximate intersection point is (2.0, 3.0).

7. A right triangle is positioned on a coordinate plane with vertices at (0,0), (12,0), and (0,5). A circle is inscribed within this triangle such that it is tangent to all three sides. What is the area of this inscribed circle? (Use π = 3.14) Answer: 12.56 Solution: Identify the triangle's side lengths. The legs are 12 units (horizontal) and 5 units (vertical). The hypotenuse can be found using the Pythagorean theorem: sqrt(12^2 + 5^2) = sqrt(144 + 25) = sqrt(169) = 13 units.
   Full step-by-step solution

   Step 1: Identify the triangle's side lengths. The legs are 12 units (horizontal) and 5 units (vertical). The hypotenuse can be found using the Pythagorean theorem: sqrt(12^2 + 5^2) = sqrt(144 + 25) = sqrt(169) = 13 units. Step 2: Calculate the area of the triangle. Area = (1/2) * base * height = (1/2) * 12 * 5 = 30 square units. Step 3: Calculate the semi-perimeter (s) of the triangle. s = (12 + 5 + 13)/2 = 30/2 = 15 units. Step 4: Use the formula for the inradius (r) of a triangle: r = Area / s = 30 / 15 = 2 units. Step 5: Calculate the area of the inscribed circle. Area = π * r^2 = 3.14 * (2)^2 = 3.14 * 4 = 12.56 square units. The answer is 12.56.