For which pair $(A,B)$,
\[
x^2+xy+y=A \\
\frac{y}{y-x}=B
\]
has no real roots?

$ \textbf{(A)}\ (1/2,2)
\qquad\textbf{(B)}\ (-1,1)
\qquad\textbf{(C)}\ (\sqrt 2, \sqrt 2)
\qquad\textbf{(D)}\ (1,1/2)
\qquad\textbf{(E)}\ (2,2/3)
$
To determine for which pair \((A, B)\) the given system of equations has no real roots, we need to analyze the discriminant of the resulting quadratic equation. The given system is:

\[
x^2 + xy + y = A
\]
\[
\frac{y}{y - x} = B
\]

First, solve the second equation for \(y\):

1. \(\frac{y}{y - x} = B\)

Multiply both sides by \(y - x\):

\[
y = B(y - x)
\]

Rearrange to isolate \(y\):

\[
y = By - Bx
\]

\[
y - By = -Bx
\]

\[
y(1 - B) = -Bx
\]

\[
y = \frac{-Bx}{1 - B}
\]

Next, substitute \(y = \frac{-Bx}{1 - B}\) into the first equation:

2. \(x^2 + x\left(\frac{-Bx}{1 - B}\right) + \frac{-Bx}{1 - B} = A\)

Simplify the equation:

\[
x^2 - \frac{Bx^2}{1 - B} - \frac{Bx}{1 - B} = A
\]

Multiply through by \(1 - B\) to clear the denominator:

\[
(1 - B)x^2 - Bx^2 - Bx = A(1 - B)
\]

Combine like terms:

\[
(1 - 2B)x^2 - Bx = A - AB
\]

Rearrange into standard quadratic form \(ax^2 + bx + c = 0\):

\[
(1 - 2B)x^2 - Bx - (A - AB) = 0
\]

Identify the coefficients \(a\), \(b\), and \(c\):

\[
a = 1 - 2B, \quad b = -B, \quad c = -A + AB
\]

For the quadratic equation to have no real roots, the discriminant must be negative. The discriminant \(\Delta\) of a quadratic equation \(ax^2 + bx + c = 0\) is given by:

\[
\Delta = b^2 - 4ac
\]

Substitute the coefficients into the discriminant formula:

\[
\Delta = (-B)^2 - 4(1 - 2B)(-A + AB)
\]

Simplify the expression:

\[
\Delta = B^2 - 4(1 - 2B)(-A + AB)
\]

\[
\Delta = B^2 - 4[(1 - 2B)(-A + AB)]
\]

\[
\Delta = B^2 - 4[-A + AB - 2AB + 2B^2A]
\]

\[
\Delta = B^2 - 4[-A + AB - 2AB + 2AB^2]
\]

\[
\Delta = B^2 - 4[-A - AB + 2AB^2]
\]

\[
\Delta = B^2 + 4A - 4AB - 8AB^2
\]

\[
\Delta = B^2 + 4A - 12AB + 8AB^2
\]

Now, we need to check which pair \((A, B)\) makes the discriminant negative. We will test each pair:

- \((A, B) = (1/2, 2)\):

\[
\Delta = 2^2 + 4 \cdot \frac{1}{2} - 12 \cdot \frac{1}{2} \cdot 2 + 8 \cdot 2^2 \]
\[
= 4 + 2 - 12 + 32 = 26 \quad (\text{positive})
\]

- \((A, B) = (-1, 1)\):

\[
\Delta = 1^2 + 4 \cdot (-1) - 12 \cdot (-1) \cdot 1 + 8 \cdot 1^2 \]
\[
= 1 - 4 + 12 + 8 = 17 \quad (\text{positive})
\]

- \((A, B) = (\sqrt{2}, \sqrt{2})\):

\[
\Delta = (\sqrt{2})^2 + 4 \cdot \sqrt{2} - 12 \cdot \sqrt{2} \cdot \sqrt{2} + 8 \cdot (\sqrt{2})^2 \]
\[
= 2 + 4\sqrt{2} - 24 + 16 = -6 + 4\sqrt{2} \quad (\text{positive})
\]

- \((A, B) = (1, 1/2)\):

\[
\Delta = \left(\frac{1}{2}\right)^2 + 4 \cdot 1 - 12 \cdot 1 \cdot \frac{1}{2} + 8 \cdot \left(\frac{1}{2}\right)^2 \]
\[
= \frac{1}{4} + 4 - 6 + 2 = \frac{1}{4} \quad (\text{positive})
\]

- \((A, B) = (2, 2/3)\):

\[
\Delta = \left(\frac{2}{3}\right)^2 + 4 \cdot 2 - 12 \cdot 2 \cdot \frac{2}{3} + 8 \cdot \left(\frac{2}{3}\right)^2 \]
\[
= \frac{4}{9} + 8 - 16 + \frac{32}{9} = \frac{36}{9} - \frac{108}{9} + \frac{32}{9} = \frac{-40}{9} \quad (\text{negative})
\]

Thus, the pair \((A, B) = (2, 2/3)\) makes the discriminant negative, indicating that the quadratic equation has no real roots.

The final answer is \( \boxed{ (2, 2/3) } \)