Solve it clear,humanized with out any Ai calculation

We are given $\sin x = \frac{8}{17}$. Since $x$ is an acute angle, $\cos x$ will be positive.
We use the Pythagorean identity $\sin^2 x + \cos^2 x = 1$.

$$\cos^2 x = 1 - \sin^2 x$$

Substitute the value of $\sin x$:

$$\cos^2 x = 1 - \left(\frac{8}{17}\right)^2$$ $$\cos^2 x = 1 - \frac{8^2}{17^2}$$ $$\cos^2 x = 1 - \frac{64}{289}$$

To subtract, find a common denominator:

$$\cos^2 x = \frac{289}{289} - \frac{64}{289}$$ $$\cos^2 x = \frac{289 - 64}{289}$$ $$\cos^2 x = \frac{225}{289}$$

Now, take the square root. Since $x$ is acute, $\cos x > 0$:

$$\cos x = \sqrt{\frac{225}{289}}$$ $$\cos x = \frac{\sqrt{225}}{\sqrt{289}}$$ $$\cos x = \frac{15}{17}$$

We are given $\sin y = \frac{5}{13}$. Since $y$ is an acute angle, $\cos y$ will be positive.
Again, use the Pythagorean identity $\sin^2 y + \cos^2 y = 1$.

$$\cos^2 y = 1 - \sin^2 y$$

Substitute the value of $\sin y$:

$$\cos^2 y = 1 - \left(\frac{5}{13}\right)^2$$ $$\cos^2 y = 1 - \frac{5^2}{13^2}$$ $$\cos^2 y = 1 - \frac{25}{169}$$

To subtract, find a common denominator:

$$\cos^2 y = \frac{169}{169} - \frac{25}{169}$$ $$\cos^2 y = \frac{169 - 25}{169}$$ $$\cos^2 y = \frac{144}{169}$$

Now, take the square root. Since $y$ is acute, $\cos y > 0$:

$$\cos y = \sqrt{\frac{144}{169}}$$ $$\cos y = \frac{\sqrt{144}}{\sqrt{169}}$$ $$\cos y = \frac{12}{13}$$

We have $\sin x = \frac{8}{17}$, $\cos y = \frac{12}{13}$, $\cos x = \frac{15}{17}$, and $\sin y = \frac{5}{13}$.

$$\frac{\sin x - \cos y}{\cos x + \sin y} = \frac{\frac{8}{17} - \frac{12}{13}}{\frac{15}{17} + \frac{5}{13}}$$

To simplify the numerator, find a common denominator for $\frac{8}{17}$ and $\frac{12}{13}$, which is $17 \times 13 = 221$:

$$\frac{8}{17} - \frac{12}{13} = \frac{8 \times 13}{17 \times 13} - \frac{12 \times 17}{13 \times 17} = \frac{104}{221} - \frac{204}{221} = \frac{104 - 204}{221} = \frac{-100}{221}$$

To simplify the denominator, find a common denominator for $\frac{15}{17}$ and $\frac{5}{13}$, which is also $17 \times 13 = 221$:

$$\frac{15}{17} + \frac{5}{13} = \frac{15 \times 13}{17 \times 13} + \frac{5 \times 17}{13 \times 17} = \frac{195}{221} + \frac{85}{221} = \frac{195 + 85}{221} = \frac{280}{221}$$

Now, substitute these back into the main expression:

$$\frac{\frac{-100}{221}}{\frac{280}{221}}$$

When dividing fractions, we multiply by the reciprocal of the denominator:

$$\frac{-100}{221} \times \frac{221}{280}$$

The 221 in the numerator and denominator cancel out:

$$\frac{-100}{280}$$

Simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 20:

$$\frac{-100 \div 20}{280 \div 20} = \frac{-5}{14}$$

The final answer is $-\frac{5}{14}$.