# Thread: find positive integers x,y,z

1. $(1)$ $\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{7}{8}(x,y\in N)$
$find$: $x,y$
$(2)$ $\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{6}{12}(x,y\in N)$
$find$: $x,y$
$(3)$ $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac {1}{z}=\dfrac{6}{24}(x,y,z\in N)$
$find$: $x,y,z$

2. Originally Posted by Albert
$(1)$ $\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{7}{8}(x,y\in N)$
$find$: $x,y$
$(2)$ $\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{6}{12}(x,y\in N)$
$find$: $x,y$
$(3)$ $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac {1}{z}=\dfrac{6}{24}(x,y,z\in N)$
$find$: $x,y,z$
without loss of generality we can take x <=y <=z as any permutation can give ans

for (1)

for (2)

without loss of generality we can take x <=y <=z as any permutation can give ans

for (1)

for (2)

4. Originally Posted by Albert

Originally Posted by Albert
$(1)$ $\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{7}{8}(x,y\in N)$
$find$: $x,y$
$(2)$ $\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{6}{12}(x,y\in N)$
$find$: $x,y$
$(3)$ $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac {1}{z}=\dfrac{6}{24}(x,y,z\in N)$
$find$: $x,y,z$
hint of (3)$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac {1}{z}=\dfrac{6}{24}$

6. Originally Posted by Albert
hint of (3)$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac {1}{z}=\dfrac{6}{24}$
The above hint is not correct because

7. Originally Posted by Albert
hint of (3)$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac {1}{z}=\dfrac{6}{24}$

9. Originally Posted by Albert
any theorem mentioning to find the amounts of solutions of this kind ,may be someone can tell me
These are known as .