1. **State the problem:** We have two real numbers $x$ and $y$ with $x \neq 0$ satisfying the equation: $$\frac{y}{x} = y^2 - 2y$$ We need to determine which of the given equalities is NOT certainly true. 2. **Given equation:** $$\frac{y}{x} = y^2 - 2y$$ 3. **Check each option:** **A.** $y^2 = 2y + \frac{y}{x}$ Start from the right side: $$2y + \frac{y}{x} = 2y + (y^2 - 2y) = y^2$$ So, A is true. **B.** $\frac{1}{x} = y - 2$ From the original equation: $$\frac{y}{x} = y^2 - 2y$$ Divide both sides by $y$ (assuming $y \neq 0$): $$\frac{\cancel{y}}{x \cancel{y}} = \frac{y^2 - 2y}{y} \Rightarrow \frac{1}{x} = y - 2$$ This matches B, so B is true if $y \neq 0$. **C.** $-\frac{y}{x} = 2y - y^2$ From the original: $$\frac{y}{x} = y^2 - 2y$$ Multiply both sides by $-1$: $$-\frac{y}{x} = - (y^2 - 2y) = 2y - y^2$$ So C is true. **D.** $y = x(y^2 - 2y)$ Multiply both sides of the original equation by $x$: $$y = x(y^2 - 2y)$$ So D is true. **E.** $\frac{2xy - xy^2 + y}{x} = 0$ Simplify numerator: $$2xy - xy^2 + y = y(2x - x y + 1)$$ Divide by $x$: $$\frac{2xy - xy^2 + y}{x} = \frac{y(2x - x y + 1)}{x} = y \frac{2x - x y + 1}{x}$$ Rewrite numerator inside fraction: $$2x - x y + 1 = x(2 - y) + 1$$ This expression is not necessarily zero for all $x,y$ satisfying the original equation. Check with original equation: From original: $$\frac{y}{x} = y^2 - 2y$$ Multiply both sides by $x$: $$y = x(y^2 - 2y)$$ Rewrite $x$: $$x = \frac{y}{y^2 - 2y}$$ Substitute into numerator: $$2x - x y + 1 = 2 \cdot \frac{y}{y^2 - 2y} - \frac{y}{y^2 - 2y} y + 1 = \frac{2y - y^2}{y^2 - 2y} + 1$$ Note that $y^2 - 2y = y(y - 2)$ and $2y - y^2 = y(2 - y)$. So numerator becomes: $$\frac{y(2 - y)}{y(y - 2)} + 1 = \frac{2 - y}{y - 2} + 1 = -1 + 1 = 0$$ Wait, this suggests numerator is zero, so the whole expression is zero. But this contradicts the earlier conclusion. Re-examining: $$\frac{2xy - xy^2 + y}{x} = \frac{y(2x - x y + 1)}{x}$$ Substitute $x = \frac{y}{y^2 - 2y}$: $$2x - x y + 1 = 2 \cdot \frac{y}{y^2 - 2y} - \frac{y}{y^2 - 2y} y + 1 = \frac{2y - y^2}{y^2 - 2y} + 1$$ Since $y^2 - 2y = y(y - 2)$ and $2y - y^2 = y(2 - y)$, then: $$\frac{2y - y^2}{y^2 - 2y} = \frac{y(2 - y)}{y(y - 2)} = \frac{2 - y}{y - 2} = -1$$ Therefore: $$-1 + 1 = 0$$ So the numerator is zero, making the whole expression zero. Hence, E is also true. 4. **Conclusion:** All equalities A, B, C, D, and E are true except possibly B if $y=0$ (division by zero in step). But the problem states $x \neq 0$ but does not restrict $y$. If $y=0$, original equation: $$\frac{0}{x} = 0^2 - 2 \cdot 0 \Rightarrow 0=0$$ True. Check B: $$\frac{1}{x} = 0 - 2 = -2$$ So $\frac{1}{x} = -2 \Rightarrow x = -\frac{1}{2}$ This is possible. So B is true for $y=0$ as well. Therefore, all options are true. But the problem says one is NOT certainly true. Re-examining B: We divided original equation by $y$ to get B, so if $y=0$, division by zero is invalid. Hence, B is not necessarily true for $y=0$. Therefore, **B is the one that is not certainly true**. **Final answer:** B