The number of pairs of integers ( $x, y$ ) satisfy $x \geq y \geq-20$ and $2 x+5 y=99$ is

We need to find pairs of integers $(x, y)$ that satisfy $2x + 5y = 99$ (the main equation) and $x \geq y \geq -20$ (the constraints).

Key insight: Since we're looking for integer solutions, we need to be systematic about finding values that work.

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From the equation $2x + 5y = 99$:

$2x = 99 - 5y$

$x = \dfrac{99 - 5y}{2}$

Important observation: For $x$ to be an integer, $(99 - 5y)$ must be even.

Since $99$ is odd, we need $5y$ to be odd, which means $y$ must be odd.

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We have two constraints on $y$:

Constraint 1: $y \geq -20$ (given directly)

Constraint 2: $x \geq y$ (from $x \geq y \geq -20$)

Substituting our expression for $x$:

$\dfrac{99 - 5y}{2} \geq y$

$99 - 5y \geq 2y$

$99 \geq 7y$

$y \leq \dfrac{99}{7} = 14.14...$

Since $y$ must be an integer: $y \leq 14$

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Combining our conditions:

$y$ must be odd (from earlier)

$-20 \leq y \leq 14$ (from constraints)

The odd integers in this range are:

$-19, -17, -15, -13, -11, -9, -7, -5, -3, -1, 1, 3, 5, 7, 9, 11, 13$

Counting these values: There are 17 possible values for $y$.

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For $y = -19$:

$x = \dfrac{99 - 5(-19)}{2} = \dfrac{99 + 95}{2} = \dfrac{194}{2} = 97$

Check: $x \geq y$? Yes, $97 \geq -19$

For $y = 13$:

$x = \dfrac{99 - 5(13)}{2} = \dfrac{99 - 65}{2} = \dfrac{34}{2} = 17$

Check: $x \geq y$? Yes, $17 \geq 13$

For $y = 15$ (checking our upper bound):

$x = \dfrac{99 - 5(15)}{2} = \dfrac{99 - 75}{2} = \dfrac{24}{2} = 12$

Check: $x \geq y$? No, $12 < 15$

This confirms our upper bound is correct.

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Since each valid $y$ value gives us exactly one corresponding $x$ value, and we found 17 valid $y$ values, there are 17 pairs of integers $(x, y)$ that satisfy the given conditions.

Answer: 17