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If a>0 and b>0, a new operation $$\nabla$$ is defined as follows:$$a \nabla b = \dfrac{a + b}{1 + ab}.$$For example,$$3 \nabla 6 = \dfrac{3 + 6}{1 + 3 \times 6} = \dfrac{9}{19}.$$For some values of x and y, the value of $$x \nabla y$$ is equal to $$\dfrac{x + y}{17}$$. How many possible ordered pairs of positive integers $x$ and $y$ are there for which this is true?

Oct 30, 2019

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Don't quite understand your question! Do you mean that for some values of a, b and for some values of x, y:

(a + b) / (1 + a*b) = (x + y) / 17 ????.

Oct 30, 2019
#2
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3 \nabla 6 = \dfrac{3 + 6}{1 + 3 \times 6} = \dfrac{9}{19}.

$$a \nabla b = \dfrac{a + b}{1 + ab}.\\ so\\ x \nabla y = \dfrac{x + y}{1 + xy}\\ but\\ x \nabla y = \dfrac{x + y}{17}\\ so\\ 1+xy=17\\~\\ xy=16\\$$

There are 5 ordered pairs of positive integers that will make this true. You can list them yourself.

Oct 31, 2019