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# help Algebra

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Given positive integers x and y such that $x\neq y$ and $\frac{1}{x} + \frac{1}{y} = \frac{1}{18} + \frac{1}{3}$, what is the smallest possible value for x + y?

Jan 30, 2022

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$$\frac{1}{x} + \frac{1}{y} = \frac{1}{18} + \frac{1}{3} = \frac{7}{18}$$

$$\frac{x+y}{xy} = \frac{7}{18}$$

$$18x + 18y = 7xy$$

$$0 = 7xy - 18x - 18y$$

$$324/7 = (7x - 18)(y-18/7)$$

$$324 = (7x - 18)(7y - 18)$$

Create a table with the factors of 324, which is what (7x - 18) and (7y - 18) will multiply to. By brute force, you can find the smallest value of x + y.

Jan 30, 2022