# What is x ,y in simultaneos equations xA2+yA2=50 y*xC2+xyC2=135

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### 1 Answer

You need to use factorial formula for `A_x^2 = (x!)/((x-2)!) ` and `A_y^2 = (y!)/((y-2)!), ` hence, converting the first equation to factorial form yields:

`(x!)/((x-2)!) + (y!)/((y-2)!) = 50`

You should convert the next equation to factorial form such that:

`y*x*(x!)/(2!(x-2)!) + x*y*(y!)/(2!(y-2)!) = 135`

You should factor out `(x*y)/(2!)` such that:

`(x*y)/(2!)*((x!)/((x-2)!) + (y!)/((y-2)!)) = 135`

Notice that you may use the first equation, hence, you may substitute 50 for `(x!)/((x-2)!) + (y!)/((y-2)!)` such that:

`(x*y)/(2!)*50 = 135`

`25xy = 135 =gt 5xy = 27 =gt xy = 27/5 =gt x = 27/(5y)`

You may return to the first equation such that:

`((x-2)!(x-1)x)/((x-2)!) + ((y-2)!(y-1)y)/((y-2)!) = 50`

`(x-1)x + (y-1)y = 50`

`x^2 - x + y^2 - y = 50`

You should remember that you may write `x^2 + y^2 = (x+y)^2 - 2xy`

`(x+y)^2 - 2xy - (x+y) = 50 `

`(x+y)^2 - (x+y) = 50 + 2xy`

`(x+y)^2 - (x+y) = 50 + 54/5`

Notice that you may factor out x+y such that:

`(x+y)(x+y-1) = (250+54)/5`

You may substitute `27/(5y)` for x such that:

`(27/(5y) + y)(27/(5y) + y - 1) = 304/5`

`5(27+5y^2)(27+5y^2 - 5y) = 304*25y^2`

`5(27+5y^2)^2 - 25y(27+5y^2) - 7600y^2 = 0`

`3645 + 1350y + 125y^2 - 675y - 125y^3 - 7600y^2 = 0`

`-125y^3 - 7475y^2 + 675y + 3645 = 0`

`-25y^3 - 1495y^2 + 135y + 729 = 0`

**You need to look the natural roots of the polynomial above and since tbis polynomial has no natural zeroes, hence, there are no natural values for x and the system has no solution.**