Expand menu menu_open Minimize Start chapters Home History history History expand_more
{{ item.displayTitle }}
navigate_next
No history yet!
Progress & Statistics equalizer Progress expand_more
Student
navigate_next
Teacher
navigate_next
{{ filterOption.label }}
{{ item.displayTitle }}
{{ item.subject.displayTitle }}
arrow_forward
No results
{{ searchError }}
search
menu_open
{{ courseTrack.displayTitle }}
{{ statistics.percent }}% Sign in to view progress
{{ printedBook.courseTrack.name }} {{ printedBook.name }}
search Use offline Tools apps
Login account_circle menu_open

Solving Linear-Quadratic Systems

Solving Linear-Quadratic Systems 1.10 - Solution

arrow_back Return to Solving Linear-Quadratic Systems
We want to solve the given system of equations using the substitution method. The variable is isolated in Equation (I). This allows us to substitute its value for in Equation (II).
Notice that in Equation (II), we have a quadratic equation in terms of only the variable. We can substitute and into the Quadratic Formula.
Simplify right-hand side
This result tells us that we have two solutions for One of them will use the positive sign and the other will use the negative sign.
Now, consider Equation (I). We can substitute and into the above equation to find the values for Let's start with
Solve for
We found that when One solution of the system, which is a point of intersection of the parabola and the line, is To find the other solution, we will substitute for in Equation (I) again.
Solve for
We found that when Therefore, our second solution, which is the other point of intersection of the parabola and the line, is