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Introduction

Introduction

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Introduction

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  1. Introduction We have worked with linear equations and equations of circles. We have solved systems of equations, including systems involving linear equations and equations of parabolas. Now we will bring together skills and concepts from these topics and apply them in this lesson to solve systems of linear equations and equations of circles. 6.3.1: Solving Systems of Linear Equations and Circles

  2. Key Concepts A system of equations is a set of equations with the same unknowns. If a two-equation system has a linear equation and a circle equation, then the system can have no real solutions, one real solution, or two real solutions. A solution is an ordered pair; its graphical representation is a point at which the line and circle intersect, as shown on the following slide. 6.3.1: Solving Systems of Linear Equations and Circles

  3. Key Concepts, continued 6.3.1: Solving Systems of Linear Equations and Circles

  4. Key Concepts, continued Systems of equations can be solved by using graphical and/or algebraic methods. The graphical method is sometimes useful only for estimating solutions. No matter which method is used to solve a system, the solution(s) should be checked algebraically by substitution. A system with two real solutions will have two points of intersection between the line and the circle. A system with one real solution will have exactly one point of intersection. 6.3.1: Solving Systems of Linear Equations and Circles

  5. Key Concepts, continued A system with no real solutions will have two complex solutions that cannot be graphed on the Cartesian coordinate plane. Remember that the standard form of an equation of a circle is (x – h)2 + (y – k)2= r2, where (h, k) is the center and r is the radius. A quadratic equation of the form ax2 + bx+ c = 0 can have no real solutions, one real solution, or two real solutions. 6.3.1: Solving Systems of Linear Equations and Circles

  6. Key Concepts, continued Using substitution, a system of a linear equation and a circle equation can be reduced to a single quadratic equation whose solutions lead to the solutions of the system. The quadratic formula states that if a quadratic equation of the form ax2 + bx+ c = 0 has one or two real solutions, then those solutions are given by 6.3.1: Solving Systems of Linear Equations and Circles

  7. Key Concepts, continued To factor a polynomial means to write it as a product of two or more polynomials. Examples: x2+ 6x + 9 = (x + 3)(x + 3) x2+ 5x – 6 = (x + 6)(x – 1) 3x2+ 10x = x(3x + 10) Note: x is a polynomial with one term, so it is a monomial. 6.3.1: Solving Systems of Linear Equations and Circles

  8. Key Concepts, continued The Zero Product Property states that if a product equals 0, then at least one of its factors is 0. Examples: If ab= 0, then a = 0 or b = 0. If (x + 6)(x – 1) = 0, then x + 6 = 0 or x – 1 = 0. Every quadratic equation of the form ax2 + bx+ c = 0 that has one or two real solutions can be solved by the quadratic formula; some can be solved more easily by factoring and using the Zero Product Property. 6.3.1: Solving Systems of Linear Equations and Circles

  9. Key Concepts, continued The distance formula states that the distance between points (x1, y1) and (x2, y2) is equal to The midpoint formula states that the midpoint of the line segment connecting points (x1, y1) and (x2, y2) is 6.3.1: Solving Systems of Linear Equations and Circles

  10. Common Errors/Misconceptions writing an incorrect sign for the last term of the trinomial that results from squaring a binomial using a instead of 2a as the denominator when applying the quadratic formula neglecting to key in parentheses when using a calculator to square a quantity neglecting to multiply both sides of an equation when multiplying to eliminate fractions forgetting to find the complex solutions to a system with no real solutions 6.3.1: Solving Systems of Linear Equations and Circles

  11. Guided Practice Example 2 Solve the system below. Check the solution(s), then graph the system on a graphing calculator. 6.3.1: Solving Systems of Linear Equations and Circles

  12. Guided Practice: Example 2, continued Solve the linear equation for one of its variables in terms of the other. 6.3.1: Solving Systems of Linear Equations and Circles

  13. Guided Practice: Example 2, continued Use the result from step 1 to substitute for y in the circle equation. 6.3.1: Solving Systems of Linear Equations and Circles

  14. Guided Practice: Example 2, continued 6.3.1: Solving Systems of Linear Equations and Circles

  15. Guided Practice: Example 2, continued Substitute the x-value from step 2 into the linear equation and find the corresponding y-value. 6.3.1: Solving Systems of Linear Equations and Circles

  16. Guided Practice: Example 2, continued The solution to the system of equations is (–4, 3). 6.3.1: Solving Systems of Linear Equations and Circles

  17. Guided Practice: Example 2, continued Check the solution by substituting it into both original equations. The solution checks. The solution of the system is (–4, 3). 6.3.1: Solving Systems of Linear Equations and Circles

  18. Guided Practice: Example 2, continued Graph the system on a graphing calculator. First, solve the circle equation for y to obtain functions that can be graphed. The linear equation solved for y is 6.3.1: Solving Systems of Linear Equations and Circles

  19. Guided Practice: Example 2, continued Graph the linear function and the two functions represented by the circle. On a TI-83/84: Step 1: Press [Y=]. Step 2: At Y1, type in [(][4][÷][3][)][X, T, θ, n][+][25] [÷][3]. Step 3: At Y2, type in [ ][25][–][X, T, θ, n][x2][)]. Step 4: At Y3, type in [(–)][ ][25][–][X, T, θ, n][x2] [)]. Step 5: Press [WINDOW] to change the viewing window. 6.3.1: Solving Systems of Linear Equations and Circles

  20. Guided Practice: Example 2, continued Step6: At Xmin, enter [(–)][12]. Step7: At Xmax, enter [12]. Step8: At Xscl, enter [1]. Step9: At Ymin, enter [(–)][8]. Step10: At Ymax, enter [8]. Step11: At Yscl, enter [1]. Step12: Press [GRAPH]. 6.3.1: Solving Systems of Linear Equations and Circles

  21. Guided Practice: Example 2, continued On a TI-Nspire: Step 1: Press the [home] key. Step 2: Navigatetothegraphsiconandpress [enter]. Step 3: Attheblinkingcursoratthebottomofthe screen, type in [(][4][÷][3][)][x][+][25][÷][3] andpress [enter]. Step 4: Move thecursorovertothe double arrow atthebottomleftofthescreenand press thecenterkeyofthenavigation pad toopen anotherequation prompt. 6.3.1: Solving Systems of Linear Equations and Circles

  22. Guided Practice: Example 2, continued Step 5: At the blinking cursor, type in [ ][25][–][x] [x2] and press [enter]. Step 6: Move the cursor over to the double arrow at the bottom left of the screen and press the center key of the navigation pad to open another equation prompt. Step 7: At the blinking cursor type in [(–)][ ][25] [–][x][x2] and press [enter]. 6.3.1: Solving Systems of Linear Equations and Circles

  23. Guided Practice: Example 2, continued ✔ 6.3.1: Solving Systems of Linear Equations and Circles

  24. Guided Practice: Example 2, continued 6.3.1: Solving Systems of Linear Equations and Circles

  25. Guided Practice Example 3 Determine whether the line with equation x = –5 intersects the circle centered at (–3, 1) with radius 4. If it does, then find the coordinates of the point(s) of intersection. 6.3.1: Solving Systems of Linear Equations and Circles

  26. Guided Practice: Example 3, continued Write the system of equations represented by the line and the circle. The equation of the circle with center (h, k) and radius r is (x – h)2 + (y – k)2 = r2, so the equation of the circle is [x – (–3)]2 + (y – 1)2 = 42, or (x + 3)2 + (y – 1)2 = 16. The system is: 6.3.1: Solving Systems of Linear Equations and Circles

  27. Guided Practice: Example 3, continued Substitute the x-value from the linear equation into the equation of the circle and then solve the resulting equation. 6.3.1: Solving Systems of Linear Equations and Circles

  28. Guided Practice: Example 3, continued Apply the quadratic formula. The solutions of ax2 + bx + c = 0 are so the solutions of an equation in the form ay2 + by+ c = 0 are 6.3.1: Solving Systems of Linear Equations and Circles

  29. Guided Practice: Example 3, continued The solutions to the quadratic equation are 6.3.1: Solving Systems of Linear Equations and Circles

  30. Guided Practice: Example 3, continued The linear equation in the system is x = –5, so any solution must have –5 as its x-coordinate. Use this fact along with the results from step 2 to write the solutions of the system. The solutions of the system are These are the coordinates of the points at which the line intersects the circle, as shown on the following graph. 6.3.1: Solving Systems of Linear Equations and Circles

  31. Guided Practice: Example 3, continued ✔ 6.3.1: Solving Systems of Linear Equations and Circles

  32. Guided Practice: Example 3, continued 6.3.1: Solving Systems of Linear Equations and Circles

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