Lesson: Solving Systems of Equations Algebraically
Example 1: Linear-Quadratic System
Solve 5x - y = 10
x2 + x - 2y = 0 Verify your solution.
Method 1: Use substitution.
Solve the linear equation for "y"
then substitute into the quadratic equation.
5x - y = 10
x2 + x - 2y = 0
5x - y = 10
x2 + x - 2y = 0
Method 2: Use elimination.
Align the terms with the same degree.
Since the quadratic term is in the variable "x",
eliminate the y-term.
5x - y = 10
x2 + x - 2y = 0
Always verify your solution...
Your Turn...
Solve the system:
3x + y = -9
4x2 - x + y = -9
Answer
Now, solve for y...
The solutions are
(0, –9) and (1, –12).
Example 2: Quadratic-Quadratic System
Using the elimination method,
eliminate "y". Solve the resulting quadratic equation for "x",
then substitute answers into either equation to solve for "y".
Solve the system:
3x2 - x - y - 2 = 0
6x2 + 4x - y = 4
Now, verify the solutions:
(-2, 12) and (1/3, -2)
8.1
Example 3: Your Turn
Answer
Solve the system and verify your solution.
2x2 + 16x + y = –26
x2 + 8x – y = –19
Solving Systems of Equations Algebraically
Workbook questions: pages 350-353