This week in Pre-Calc we learned about Solving Quadratic Inequalities in One Variable, Graphing Linear Inequalities in Two Variables, and Solving Quadratic Systems of Equations.
When solving Quadratic Systems of Equations you would solve algebraically, either with substitution, or elimination. You could also graph, but if your point of intersection is not on a prefect point, you will not be able to be completely accurate. To solve get everything equal to zero.
- y = x2 – 5x + 7
- y = 2x + 1
Make both equations into “y=” format:
They are both in “y=” format, so Set them equal to each other
Simplify into “= 0” format (like a standard Quadratic Equation)
Which gives us the solutions x=1 and x=6
Use the linear equation to calculate matching “y” values, so we get (x,y) points as answers
The matching y values are:
- for x=1: y = 2x+1 = 3
- for x=6: y = 2x+1 = 13
Our solution: the two points are (1,3) and (6,13)