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Introduction Previously, we learned how to solve quadratic-linear systems by graphing and identifying points of intersection. In this lesson, we will focus.

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Presentation on theme: "Introduction Previously, we learned how to solve quadratic-linear systems by graphing and identifying points of intersection. In this lesson, we will focus."— Presentation transcript:

1 Introduction Previously, we learned how to solve quadratic-linear systems by graphing and identifying points of intersection. In this lesson, we will focus on solving a quadratic-linear system algebraically. When doing so, substitution is often the best choice. Substitution is the replacement of a term of an equation by another that is known to have the same value. 1 5.4.2: Solving Systems Algebraically

2 Key Concepts When solving a quadratic-linear system, if both functions are written in function form such as “y =” or “f(x) =”, set the equations equal to each other. When you set the equations equal to each other, you are replacing y in each equation with an equivalent expression, thus using the substitution method. 2 5.4.2: Solving Systems Algebraically

3 Key Concepts You can solve by factoring the equation or by using the quadratic formula, a formula that states the solutions of a quadratic equation of the form ax 2 + bx + c = 0 are given by 3 5.4.2: Solving Systems Algebraically

4 Common Errors/Misconceptions miscalculating signs incorrectly distributing coefficients 4 5.4.2: Solving Systems Algebraically

5 Guided Practice Example 1 Solve the given system of equations algebraically. 5 5.4.2: Solving Systems Algebraically

6 Guided Practice: Example 1, continued 1.Since both equations are equal to y, substitute by setting the equations equal to each other. 6 5.4.2: Solving Systems Algebraically –3x + 12 = x 2 – 11x + 28Substitute –3x + 12 for y in the first equation.

7 Guided Practice: Example 1, continued 2.Solve the equation either by factoring or by using the quadratic formula. Since a (the coefficient of the squared term) is 1, it’s simplest to solve by factoring. 7 5.4.2: Solving Systems Algebraically –3x + 12 = x 2 – 11x + 28 Equation 0 = x 2 – 8x + 16 Set the equation equal to 0 by adding 3x to both sides, and subtracting 12 from both sides. 0 = (x – 4) 2 Factor.

8 Guided Practice: Example 1, continued Substitute the value of x into the second equation of the system to find the corresponding y-value. For x = 4, y = 0. Therefore, (4, 0) is the solution. 8 5.4.2: Solving Systems Algebraically x – 4 = 0 Set each factor equal to 0 and solve. x = 4 y = –3(4) + 12Substitute 4 for x. y = 0

9 Guided Practice: Example 1, continued 3.Check your solution(s) by graphing. 9 5.4.2: Solving Systems Algebraically

10 Guided Practice: Example 1, continued The equations do indeed intersect at (4, 0); therefore, (4, 0) checks out as the solution to this system. 10 5.4.2: Solving Systems Algebraically ✔

11 Guided Practice: Example 1, continued 11 5.4.2: Solving Systems Algebraically

12 Guided Practice Example 2 Solve the given system of equations algebraically. 12 5.4.2: Solving Systems Algebraically

13 Guided Practice: Example 2, continued 1.Since both equations are equal to y, substitute by setting the equations equal to each other. 13 5.4.2: Solving Systems Algebraically x – 1 = 2x 2 + 13x + 15 Substitute x – 1 for y in the first equation.

14 Guided Practice: Example 2, continued 2.Solve the equation either by factoring or by using the quadratic formula. Since the equation can be factored easily, choose this method. 14 5.4.2: Solving Systems Algebraically x – 1 = 2x 2 + 13x + 15Equation 0 = 2x 2 + 12x + 16 Set the equation equal to 0 by subtracting x from both sides and then adding 1 to both sides. 0 = 2(x + 2)(x + 4)Factor.

15 Guided Practice: Example 2, continued Next, set the factors equal to 0 and solve. Substitute each of the values you found for x into the second equation of the system to find the corresponding y-value. 15 5.4.2: Solving Systems Algebraically x + 2 = 0x + 4 = 0 x = –2x = –4 y = (–2) – 1Substitute –2 for x. y = –3

16 Guided Practice: Example 2, continued For x = –2, y = –3, and for x = –4, y = –5. Therefore, (–2, –3) and (–4, –5) are the solutions to the system. 16 5.4.2: Solving Systems Algebraically y = (–4) – 1Substitute –4 for x. y = –5

17 Guided Practice: Example 2, continued 3.Check your solution(s) by graphing. 17 5.4.2: Solving Systems Algebraically

18 Guided Practice: Example 2, continued The equations do indeed intersect at (–2, –3) and (–4, –5); therefore, (–2, –3) and (–4, –5) check out as the solutions to this system. 18 5.4.2: Solving Systems Algebraically ✔

19 Guided Practice: Example 2, continued 19 5.4.2: Solving Systems Algebraically

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