1. 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

2. 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.

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

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

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

6. 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!

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

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

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

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

11. 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!

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

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

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

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

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

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

18. 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!

19. 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:

20. 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: