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:

#4x^2-5y^2=16" [1]"#
#3x+y=6" [2]"#

Write equation [2] as y in terms of x:

#y=6-3x" [2.1]"#

Substitute the right side of equation [2.1] for y in equation [1]:

#4x^2-5(6-3x)^2=16#

Expand the square:

#4x^2-5(9x^2 - 36x + 36)=16#

Distribute the -5:

#4x^2- 45x^2 +180x -180=16#

Combine like terms:

#-41x^2 +180x -196=0#

Multiply both sides by -1:

#41x^2 -180x +196=0#

Factor:

#(x-2)(41x-98) = 0#

#x = 2 and x = 98/41#

Use equation [2.1] to find the corresponding values of y:

#y = 6-3(2) = 0# and y = 6 - 3(98/41) = -48/41

The two points are: #(2,0) and (98/41,-48/41)#

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]}