The expression \( -8 a x y+\frac{7 a^{2} y}{5} \) can be written in the form \( \frac{h a y}{5}(7 a+k x) \). Find the values of \( h \) and \( k \).

To rewrite the expression \( -8 a x y + \frac{7 a^2 y}{5} \) in the form \( \frac{h a y}{5}(7 a + k x) \), we start by factoring out common terms. First, we notice that both terms in the original expression share a common factor of \( a y \). We can factor \( a y \) out: \[ -8 a x y + \frac{7 a^2 y}{5} = a y \left(-8 x + \frac{7 a}{5}\right) \] Next, we want to express this in the desired form \( \frac{h a y}{5}(7 a + k x) \). To do this, we can rewrite the expression inside the parentheses: \[ -8 x + \frac{7 a}{5} = \frac{-40 x + 7 a}{5} \] Now, we can substitute this back into our expression: \[ a y \left(-8 x + \frac{7 a}{5}\right) = a y \cdot \frac{-40 x + 7 a}{5} \] This can be rewritten as: \[ \frac{a y}{5}(-40 x + 7 a) \] Now, we can compare this with the desired form \( \frac{h a y}{5}(7 a + k x) \). We can rearrange \( -40 x + 7 a \) to match the form: \[ -40 x + 7 a = 7 a - 40 x \] Thus, we can identify: \[ h = 1, \quad k = -40 \] Therefore, the values of \( h \) and \( k \) are: \[ \boxed{1} \quad \text{and} \quad \boxed{-40} \]