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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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 : Solving Systems Algebraically

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 : Solving Systems Algebraically

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 : Solving Systems Algebraically

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

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

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

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 : 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.

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 : 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

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

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

Guided Practice: Example 1, continued : Solving Systems Algebraically

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

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

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 : Solving Systems Algebraically x – 1 = 2x x + 15Equation 0 = 2x x + 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.

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 : Solving Systems Algebraically x + 2 = 0x + 4 = 0 x = –2x = –4 y = (–2) – 1Substitute –2 for x. y = –3

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 : Solving Systems Algebraically y = (–4) – 1Substitute –4 for x. y = –5

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

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 : Solving Systems Algebraically ✔

Guided Practice: Example 2, continued : Solving Systems Algebraically