# 1)Solve the equations for real x and y:5x{1 + 1/(x^2+y^2)}=125y(1 - 1/(x^2+y^2)}=42)Solve for real x and y:xy^2 = 15x^2 + 17xy + 15y^2x^2y = 20x^2 + 3y^2

mlehuzzah | Certified Educator

1)

`5x(1+(1)/(x^2+y^2))=12`

`1+(1)/(x^2+y^2) = (12)/(5x)`

`(1)/(x^2+y^2) = (12)/(5x)-1`

`5y(1-(1)/(x^2+y^2))=4`

`1-(1)/(x^2+y^2)=(4)/(5y)`

`-(1)/(x^2+y^2)=(4)/(5y) - 1`

`(1)/(x^2+y^2)=-(4)/(5y) + 1`

So:

`(12)/(5x)-1 = -(4)/(5y) + 1`

`(12)/(5x) = -(4)/(5y) + 2` (added 1 to both sides)

`(12y)/(x) = -(4) + 2 (5y)` (multiplied both sides by 5y)

`12y = -4x + 10xy` (multiplied both sides by x)

`4x = 10xy - 12y`

`4x = y (10x - 12)`

`y = (4x)/(10x - 12)

`y = (2x)/(5x - 6)

So, for any number x, except `x != (6)/(5)` we have a pair of real numbers:

`x, (2x)/(5x - 6)` that solves the equations.

BUT: we notice that in the original equation, there is `(1)/(x^2+y^2)`

We can't divide by 0, so we also are not allowed to include the solution `(0,0)`

On a graph, the set of points is:

2)

`xy^2 = 15x^2 + 17xy + 15y^2`

`x^2 y^2 = 15x^3 + 17 x^2 y + 15 x y^2`

`x^2y = 20x^2 + 3y^2 `

`x^2 y^2 = 20 x^2 y + 3 y^3`

So:

`15x^3 + 17 x^2 y + 15 x y^2 = 20 x^2 y + 3 y^3`

`15x^3 + 15 x y^2 = 3 x^2 y + 3 y^3`

`5x^3 + 5 x y^2 = x^2 y + y^3`

`5x (x^2 + y^2) = y (x^2 + y^2)`

If `x^2 + y^2 = 0`, then both sides are 0, and `x=y=0` is the solution. If `x^2 + y^2 != 0`, then we can divide both sides by it:

`5x = y`

This set of points is a line: `(x,5x)`