1. Solving Equations and Inequalities with Technology
1) The first few slides are intended to “set the stage” for the concepts being presented. Specifically:
An equation in one variable can be thought of graphically as two functions; “y” equal to the left side and “y” equal to the right side. The solution is the x-coordinate of the point of intersection;
An inequality in one variable can be thought of graphically as two functions where each side represents a function, such that the left side is either “above” or “below” the right side. The solution still depends on knowing the x-coordinate of the point of intersection to determine the end-point(s) of the required interval;
In some cases it is necessary to access technology to determine both the graph of the two functions and the point of intersection.
2) The content, although intended for Unit 6 expectation D3.2, can be used to explore equations and inequalities in other strands of the course.
3) Solutions to examples advance in algebraic increments. That is, the early examples (linear, quadratic and rational equations and linear inequality) can be solved algebraically with relative ease. The quadratic and rational inequality require students to use cases in the algebraic solution.
The last 2 equations and corresponding inequalities can only be solved with the aid of technology.

2. Solve: NOW, consider TWO functions y = Left Side and y = Right Side
y = 5x – and y = 2
Notes to Teacher:

3. x = 1

4. x = 1

5. Blue Graph ABOVE Red Graph
NOTE
The y-value of the function corresponding to the Left Side is greater than the y-value of the function corresponding to the Right Side
The function corresponding to the Left Side is above the function corresponding to the Right Side
Blue Graph ABOVE Red Graph
x < 1

6. Solve:
x = -2
x = 3

7. For what values of x is the quadratic is ABOVE the linear?
Now consider . . .
x = -2
x = 3
x < -2 OR
x > 3
For what values of x is the quadratic is ABOVE the linear?

8. Solve:

9. Consider the inequality:
Red Below Blue
…which is the solution to:
A “small” gap for -1 < x < - 0.8
x = - 0.8
x ≤ -1 or
x ≥ - 0.8

10. Using technology, the intersection points will be . . .
All of the early examples COULD be solved algebraically. Now consider
(1.73, 0.99)
x = OR
x = 1.73
(-1.06, -0.87)

11. Consider the solution to the corresponding inequality.
What is the solution for: x2 > sin(x)?
(1.73, 0.99)
< x < 1.73
(-1.06, -0.87)

12. Now consider the solution to:
(3.54, 4.21)
(-0.30, 0.88)
(-2.95, 0.30)
x = 3.54
x = -2.95 OR
x = -0.30 OR

13. And the inequality: (3.54, 4.21) (-0.30, 0.88) (-2.95, 0.30)
-2.95 ≤ x ≤
x ≥ 3.54

14. and there’s much, much more . . .