Question

How do you solve the system `4x^2-5y^2=16` and `3x+y=6`?

Answer 1

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]}