Prove the for every real number x, there exists a y such that..?

Question

(x+y = xy) iff x is not equal to 1.

I need a systematic proof (step-by-step). Thanks a whole bunch!

Answer 1

x + y = xy ==> x = xy - y ==> x = y (x - 1) ==> y = x / (x - 1). So y evaluates to a real value for any x except x = 1, because x = 1 gives a denominator of x - 1 = 1 - 1 = 0, and we cannot divide by zero.

Answer 2

For the 2d assume that F(x) is a polynomial with optimal skill n. F(F(x)) will comprise skill 2n and so the equation F(F(x)) = 6x - F(x) can't be an identity different than n = a million. enable F(x) = ax + b ----> F(F(x)) = a(ax + b) + b = (a^2)x + ab + b F(F(x)) = 6x - F(x) ----> (a^2)x + ab + b = 6x - ax - b (a^2 + a - 6)x + (a + 2)b = 0 For this to be an identity we want a^2 + a - 6 = 0 and (a + 2)b = 0 the 1st factors a = 2 or -3, the 2d factors a = -2 or b = 0. y = -3x does no longer be effective for effective x so a = 2, b = 0 is the sole answer. So y = 2x is the sole answer IF F(x) is a polynomial. i think of of that by connection with skill series all distinctive purposes are eradicated yet i'm unsure.

Answer 3

x + y = xy so x = xy -y
x = y(x-1)
x/(x-1) =y
y is defined for all x not= 1
that proves A(xnot=1) E(y) suchthat (x+y =xy) since we have a well formed formula for computing the y A(xnot=1)
Now the other way:
x+y=xy implies the y=x/(x-1) formula.
so we have iff.

Answer 4

1) adding (- x - xy) to both members
y - xy = -x

2) multiplying by (-1) both members
xy - y = x

3) collecting y on the left member
(x - 1)y = x

4) dividing per (x - 1) both members [being x not equal to 1, you can divide every number by (x - 1)
y = x/(x - 1)

So the number y you looked for is x/(x - 1).