# simultaneous equations `8A_x^y=A_x^(y+1) ; 8C_x^y=5C_x^(y+1)` x?y?

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You should solve the following system of simultaneous equations such that:

`{(8A_x^y = A_x^(y+1)),(8C_x^y = 5C_x^(y+1)):}`

You need to use factorial formula such that:

`C_x^y = (A_x^y)/(P_y) ; C_x^(y+1) = (A_x^(y+1))/(P_(y+1))`

`{(8A_x^y = A_x^(y+1)),(8((A_x^y)/(P_y)) = 5((A_x^(y+1))/(P_(y+1)))):}`

Notice that you may substitute `A_x^(y+1)` for `8A_x^y` such that:

`{(8A_x^y = A_x^(y+1)),((A_x^(y+1)/(P_y)) = 5((A_x^(y+1))/(P_(y+1)))):}`

Reducing like terms yields:

`{(8A_x^y = A_x^(y+1)),(1/(P_y) = 5/(P_(y+1))):}`

Solving the second equation yields:

`5P_y = P_(y+1)`

You should remember that `P_y = y! ` and P_(y+1) = (y+1)! = y!(y+1) such that:

`5y! = y!(y+1) => 5 = y + 1 => y = 4`

You need to substitute 4 for y in the first equation such that:

`8A_x^4 = A_x^5 => (8x!)/((x-4)!) = (x!)/((x-5)!)`

Notice that `(x-4)! = (x-5)!(x-4)` such that:

`8/((x-5)!(x-4)) = 1/((x-5)!) => 8/(x-4) = 1 => 8 = x-4 => x = 12`

**Hence, evaluating the solution to the system of simultaneous equations yields `x = 12, y = 4` .**