Implicit Differentiation

Woodse07

Could anyone help me try to solve these equations?

Solve the system
x^2+xy=1,xy+y^2=8.

AngryHippie

I don't think you need to drag calculus into it.

If you sum those two equations, you get x^2+y^2+2xy=9.

This has two real roots, either x+y, or x-y. giving the answers to be 3, or -3

The original equations neatly turn out to be x(x+y)=1 and y(x+y)=8.

Substituting in the values of 3, or -3, thus solving for x we get 3x=1, or -3x=1 => x=+/- 1/3
Substituting in the values of 3, or -3, thus solving for y we get 3y=8, or -3y=8
=> y= +/- 3/8


On second look, the negative solutions are not valid. The answer is x = 1/3 and y=8/3
Checking against the original equations proves the solution to be valid.

Mwalimu

Adding the two given equations produces x^2 + 2xy + y^2 = 9, as noted by AngryHippie. This gives a perfect square (x+y)^2 = 9, so that (x+y) = +/- 3 (note that (x-y) does not arise here).

Dividing the two given equations (assuming non-zero denominator) produces 8x = y.
Hence, substituting for y yields 9x = 3 or 9x = -3 and thus x = +1/3 or -1/3, with corresponding values of y being +8/3 or -8/3.

Hence, the solution to the original pair of equations is either [x = 1/3, y = 8/3]
or [x = -1/3, y = -8/3].

Substitution into the original equations confirms both solutions.