Product systems that compare numbers. Game on.

Let $x$ be the first integer and $y$ the second. Then $xy = 260$ and $x = 2y + 6$. From the product, $y = \dfrac{260}{x}$. Substitute into the second equation: $x = 2\left(\dfrac{260}{x}\right) + 6$. Multiply through by $x$ to clear the denominator: $x^{2} = 520 + 6x$. Rearrange: $x^{2} - 6x - 520 = 0$. Factor: $(x - 26)(x + 20) = 0$. The positive solution is $x = 26$, so $y = \dfrac{260}{26} = 10$. The smaller integer is $10\textsf{.}$

Let $x$ be the first integer and $y$ the second. Then $xy = 156$ and $x = 4y + 2$. From the product, $y = \dfrac{156}{x}$. Substitute into the second equation: $x = 4\left(\dfrac{156}{x}\right) + 2$. Multiply through by $x$ to clear the denominator: $x^{2} = 624 + 2x$. Rearrange: $x^{2} - 2x - 624 = 0$. Factor: $(x - 26)(x + 24) = 0$. The positive solution is $x = 26$, so $y = \dfrac{156}{26} = 6$. The smaller integer is $6\textsf{.}$

Let $x$ be the first integer and $y$ the second. Then $xy = 324$ and $x = 2y + 3$. From the product, $y = \dfrac{324}{x}$. Substitute into the second equation: $x = 2\left(\dfrac{324}{x}\right) + 3$. Multiply through by $x$ to clear the denominator: $x^{2} = 648 + 3x$. Rearrange: $x^{2} - 3x - 648 = 0$. Factor: $(x - 27)(x + 24) = 0$. The positive solution is $x = 27$, so $y = \dfrac{324}{27} = 12$. The smaller integer is $12\textsf{.}$

Let $x$ be the first integer and $y$ the second. Then $xy = 288$ and $x = 3y + 5$. From the product, $y = \dfrac{288}{x}$. Substitute into the second equation: $x = 3\left(\dfrac{288}{x}\right) + 5$. Multiply through by $x$ to clear the denominator: $x^{2} = 864 + 5x$. Rearrange: $x^{2} - 5x - 864 = 0$. Factor: $(x - 32)(x + 27) = 0$. The positive solution is $x = 32$, so $y = \dfrac{288}{32} = 9$. The larger integer is $32\textsf{.}$

Let $x$ be the first integer and $y$ the second. Then $xy = 266$ and $x = 5y + 3$. From the product, $y = \dfrac{266}{x}$. Substitute into the second equation: $x = 5\left(\dfrac{266}{x}\right) + 3$. Multiply through by $x$ to clear the denominator: $x^{2} = 1330 + 3x$. Rearrange: $x^{2} - 3x - 1330 = 0$. Factor: $(x - 38)(x + 35) = 0$. The positive solution is $x = 38$, so $y = \dfrac{266}{38} = 7$. The smaller integer is $7\textsf{.}$

Let $x$ be the first integer and $y$ the second. Then $xy = 510$ and $x = 2y + 4$. From the product, $y = \dfrac{510}{x}$. Substitute into the second equation: $x = 2\left(\dfrac{510}{x}\right) + 4$. Multiply through by $x$ to clear the denominator: $x^{2} = 1020 + 4x$. Rearrange: $x^{2} - 4x - 1020 = 0$. Factor: $(x - 34)(x + 30) = 0$. The positive solution is $x = 34$, so $y = \dfrac{510}{34} = 15$. The smaller integer is $15\textsf{.}$

Let $x$ be the first integer and $y$ the second. Then $xy = 385$ and $x = 3y + 2$. From the product, $y = \dfrac{385}{x}$. Substitute into the second equation: $x = 3\left(\dfrac{385}{x}\right) + 2$. Multiply through by $x$ to clear the denominator: $x^{2} = 1155 + 2x$. Rearrange: $x^{2} - 2x - 1155 = 0$. Factor: $(x - 35)(x + 33) = 0$. The positive solution is $x = 35$, so $y = \dfrac{385}{35} = 11$. The larger integer is $35\textsf{.}$

Let $x$ be the first integer and $y$ the second. Then $xy = 130$ and $x = 4y + 6$. From the product, $y = \dfrac{130}{x}$. Substitute into the second equation: $x = 4\left(\dfrac{130}{x}\right) + 6$. Multiply through by $x$ to clear the denominator: $x^{2} = 520 + 6x$. Rearrange: $x^{2} - 6x - 520 = 0$. Factor: $(x - 26)(x + 20) = 0$. The positive solution is $x = 26$, so $y = \dfrac{130}{26} = 5$. The smaller integer is $5\textsf{.}$