Question

How do you solve the system `4x^2-56x+9y^2+160=0` and `4x^2+y^2-64=0`?

Answer 1

`y^2 = 64 - 4x^2`

`y =+- sqrt(64 - 4x^2)`

`4x^2 - 56x + 9(sqrt(64 - 4x^2))^2 + 160 = 0`

`4x^2 - 56x + 9(64 - 4x^2) + 160 = 0`

`4x^2 - 56x + 576 - 36x^2 + 160 = 0`

`0 = 32x^2 + 56x - 736`

`0 = 8(4x^2 + 7x - 92)`

`0 = 8(4x^2 - 16x + 23x - 92)`

`0 = 8(4x(x - 4) + 23(x - 4))`

`0 = 8(4x + 23)(x - 4)`

`x = -23/4 and 4`

Case 1:

`4(-23/4)^2 + y^2 - 64 = 0`

`4(529/16) + y^2 - 64 = 0`

`y^2 = 64 - 529/4`

`y = O/`

Case 2:

`4(4)^2+ y^2 - 64 = 0`

`4(16) + y^2 - 64 = 0`

`y^2 = 64 - 64`

`y^2 = 0`

`y = 0`

The only real solution is `x = 4`, `y = 0`.

Thus, the solution set is `{-4, 0}`.

Hopefully this helps!

Answer 2

`y = 0` and `x = 4`

Explanation:

Making `x_2=x^2,y_2=y^2` and solving for `x_2,y_2` the system

#{ (4 x_2+9y_2=56x-160), (4x_2+y_2=64) :}#

we obtain

# { (x_2 = 1/4 (92 - 7 x)), (y_2 = 7 x-28) :}#

Now we solve

`x^2 = 1/4 (92 - 7 x)`

obtaining `x = {-23/4,4}`

and

`y = pm sqrt(7x-28)`

or

`y = pm sqrt(7 cdot 4-28) = 0` excluding complex solutions.

we have

`y = 0` and `x = 4`