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Introduction

Introduction

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Introduction

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  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. 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. 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 ax2+ bx + c = 0 are given by 5.4.2: Solving Systems Algebraically

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

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

  6. Guided Practice: Example 1, continued Since both equations are equal to y, substitute by setting the equations equal to each other. 5.4.2: Solving Systems Algebraically

  7. Guided Practice: Example 1, continued 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. 5.4.2: Solving Systems Algebraically

  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. 5.4.2: Solving Systems Algebraically

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

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

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

  13. Guided Practice: Example 2, continued Since both equations are equal to y, substitute by setting the equations equal to each other. 5.4.2: Solving Systems Algebraically

  14. Guided Practice: Example 2, continued Solve the equation either by factoring or by using the quadratic formula. Since the equation can be factored easily, choose this method. 5.4.2: Solving Systems Algebraically

  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. 5.4.2: Solving Systems Algebraically

  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. 5.4.2: Solving Systems Algebraically

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

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

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