Warm Up Solve each quadratic equation by factoring. Check your answer. 1. x2 - 3x - 10 = 0 5, -2 2. -3x2 - 12x = 12 -2 Find the number of real solutions of each equation using the discriminant. 3. 25x2 - 10x + 1 = 0 one 4. 2x2 + 7x + 2 = 0 two 5. 3x2 + x + 2 = 0 none

Recall that a system of linear equations is a set of two or more linear equations. A solution of a system is an ordered pair that satisfies each equation in the system. Points where the graphs of the equations intersect represent solutions of the system. A nonlinear system of equations is a system in which at least one of the equations is nonlinear. For example, a system that contains one quadratic equation and one linear equation is a nonlinear system.

A system made up of a linear equation and a quadratic equation can have no solution, one solution, or two solutions, as shown below.

Example 1: Solving a Nonlinear System by Graphing Solve the system by graphing. Check your answer. y = x2 + 4x + 3 y = x + 3

Check It Out! Example 1 1. Solve the system by graphing. Check your answer. y = x2 - 4x + 5 y = x + 1

The substitution method is a good choice when either equation is solved for a variable, both equations are solved for the same variable, or a variable in either equation has a coefficient of 1 or -1. Remember!

Example 2: Solving a Nonlinear system by substitution. Solve the system by substitution. y = x2 - x - 5 y = -3x + 3

Example 2: Continued The solutions are (4, 15) and (2, –3).

Check It Out! Example 2 1. Solve the system by substitution. Check your answer. y = 3x2 - 3x + 1 y = -3x + 4

Check It Out! Example 2 Continued The solutions are (–1, 7) and (1, 1).

The elimination method is a good choice when both equations have the same variable term with the same or opposite coefficients or when a variable term in one equation is a multiple of the corresponding variable term in the other equation. Remember!

Example 3 : Solving a Nonlinear System by Elimination. Solve each system by elimination. 3x - y = 1 y = x2 + 4x - 7 A

Example 3 : Continued The solution is (–3, –10 ) and (2, 5).

Example 3 : Continued y = 2x2 + x - 1 x - 2y = 6 B

Example 3 : Continued x= - 1 ± √ – 63 8 Since the discriminant is negative, there are no real solutions

Check It Out! Example 3 1. Solve each system by elimination. Check your answers.. 2x - y = 2 y = x2 - 5 a

Check It Out! Example 3 Continued The solution is (3, 4) and (–1, –4).

The elimination method is a good choice when both equations have the same variable term with the same or opposite coefficients or when a variable term in one equation is a multiple of the corresponding variable term in the other equation. Remember!

Lesson Quiz: Part-1 Solve each system by the indicated method. y = x2 - 4x + 3 y = x - 1 (1, 0), (4, 3) 1. Graphing: y = 2x2 - 9x - 5 y = -3x + 3 (-1, 6), (4, -9) 2. Substitution:

Lesson Quiz: Part-2 y = x2 + 2x - 3 x - y = 5 no solution 3. Elimination: y = x2 - 7x + 10 2x - y = 8 (3, -2), (6, 4) 4. Elimination: