1. Bell Work: Simplify: (0.04 x 10 )(50 x 10 ) (0.000004)(50,000) -9 16

2. Answer: 1 x 10 8

3. Lesson 81: Graphical Solutions, Inconsistent Equations, Dependent Equations

4. In lesson 48, we noted that the degree of a term of a polynomial is the sum of the exponents of the variables in the term. Thus 2xy is a sixth-degree term, 4x is a first-degree term, and xy is a second-degree term. 5

5. Also, we remember the the degree of a polynomial is the same as the degree of its highest-degree term and that the degree of a polynomial equation is the same as the degree of the highest-degree in the equation. Thus x y + 4y is a 5 th degree polynomial. x y + 4y = x is a 5 th degree polynomial equation. x + 2y = 4 is a 1 st degree polynomial equation. 4444

6. We have learned to find the solution to a system of two linear equations by using the substitution method and elimination method. Today we will see that we can find the solution to a system of two linear equations by graphing each of the equations and visually estimating the coordinates of the point where the two lines cross.

7. The downside of this method is that it is an inexact method because the coordinates of the crossing point must be estimated. We can check to see whether the coordinates of the crossing point are really the solutions of both equations by checking to see if the coordinates satisfy both equations.

8. Example: Solve by graphing: y = x + 1 y = -2x + 4

9. Answer: It appears that the lines cross at x = 1 and y = 2. 2 = 1 + 1 2 = -2(1) + 4

10. Example: Solve by graphing: y = 2 y = x

11. Answer: (2, 2)

12. Example: Solve by graphing: y = x + 2 y = x – 1

13. Answer: No solution We call these inconsistent equations

14. Inconsistent equations are equations that have no common solution.

15. We have discussed substitution and elimination in great detail. We saw that before either method was used, an assumption was necessary. We assumed that a value of x and y existed that would satisfy both equations. Seeing the previous examples we see this assumption was invalid. There is no point that is common.

16. Example: Determine whether the following set of equations have a common solution: y = x + 2 y = x – 1 (hint: use substitution)

17. Answer: x – 1 = x + 2 -1 = 2 False These equations do not share a common solution.

18. We have defined equivalent equation to be equations that have the same solution set. If we multiply every term on both sides of an equation by the same nonzero quantity, the resulting equation is an equivalent equation to the original equation.

19. For example, graph the equation y – x = 2. Now if we multiply every term in the equation by some quantity, say the number 2, we get the equation 2y – 2x = 4, which is an equivalent equation to the original equation. Thus, all ordered pairs of x and y that satisfy the original equation also satisfy the new equation, and the graphs of the equations are the same.

20. This means that any ordered pair that is a solution to one of the equations is also a solution to the other equation. Equivalent linear equations are called dependent equations.

21. Example: Determine whether the following set of equations is dependent: y – x = 2 2y – 2x = 4

22. Answer: Use either substitution or elimination. The equations are dependent.

23. Summary: Two linear equations in two unknowns fall into one of three categories: 1.Consistent Equations 2.Inconsistent Equations 3.Dependent Equations

25. Example: Consider the equations y = 2x + 5 and y = x. is this pair of equations consistent, inconsistent, or dependent?

26. Answer: Use substitution: x = 2x + 5 x = -5 (-5, -5) Consistent

27. HW: Lesson 81 #1-30