If x does not = 0, and xy+1=(x+1)(y+1), what does y/x+1=?

Question

The answer is 0, but I don't know why...

Answer 1

xy+1=(x+1)(y+1)

xy + 1 = xy + x + y + 1

x + y = 0

x = -y

y/x = y/-y = -1

-1 + 1 = 0

Done.

Answer 2

If xy+1 = (x+1)(y+1)...
xy+1 = xy + x + y + 1
0 = x+y
-x = y
-1 = y/x

So...
y/x + 1 = -1 + 1 = 0

Answer 3

xy+1 = (x+1)(y+1)
xy+1 = xy+y+x+1
0 = xy+y+x+1-xy-1
0 = y+x
y+x = 0

devide both sides by x

y/x + x/x = 0/x

y/x + 1 = 0

Answer 4

Here is the answer:-

xy+1= (x+1) (y+1)
xy= (x+1) (y+1) -1
Dividing b/s by x^2, we get,

y/x = (x+1) (y+1) / x^2 - 1 / x^2

Now adding +1 b/s, we get,
y/x +1= (x+1) (y+1) / x^2 - 1 / x^2 +1
Solving the above equation we get,
y/x + 1 = y/x +1/x + y / x^2 +1
On b/s, we have y/x commom and 1 is also common, therefore they will be cancelled and we get,

y/x^2 = -1/x
Multiplying b/s by x, we get,
y/x = -1
Now, adding b/s +1, we get,
y/x +1 = -1+1
y/x +1 = 0, which is the answer of the equation.

Answer 5

xy+1= xy+x+y+1
x+y=0
Dividing throughout by x
y/x+1=0

Answer 6

solvin the second equation it is that:
xy + 1 = xy + x + y +1
hence
x = - y
now
(y/x) + 1 = (y + x)/ x
but x = - y
hence
(y + x)/ x = (y - y)/x = 0/x = 0

This all is true if the 3rd equation is (y/x) + 1 =?

Answer 7

xy+1=(x+1)(y+1)
multiply the left terms
xy+1 = xy +y +x +1
simplify
0 = x + y
so x = -y
Now substitute in this equation
y/x +1 = ?
y/-y +1 = ?
y/-y is -1
-1 +1 = 0