How do you solve the system #4x^2-5y^2=16# and #3x+y=6#?
1 Answer
Oct 14, 2017
For this type of system, I use the substitution method.
Explanation:
Given:
Write equation [2] as y in terms of x:
Substitute the right side of equation [2.1] for y in equation [1]:
Expand the square:
Distribute the -5:
Combine like terms:
Multiply both sides by -1:
Factor:
Use equation [2.1] to find the corresponding values of y:
The two points are:
You may find these two points of intersection on the following graph:
graph{(4x^2-5y^2-16)(3x+y-6)=0 [-10, 10, -5, 5]}