# I am having trouble finding all solutions (or at least proving I have all the solutions already). The equation is 1/a+1/b+2/c=1 *a,b,c are positive I tried to base it on the solutions of a similar equation with the last term a 1 instead of a 2.

Integer solutions to fraction equation
I am having trouble finding all solutions (or at least proving I have all the solutions already).
The equation is
$\frac{1}{a}+\frac{1}{b}+\frac{2}{c}=1$
*a,b,c are positive I tried to base it on the solutions of a similar equation with the last term a 1 instead of a 2.
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Let $a\ge b$
Hence,
$1=\frac{1}{a}+\frac{1}{b}+\frac{2}{c}\le \frac{2}{b}+\frac{2}{c},$
which gives
$bc\le 2b+2c$
or
$\left(b-2\right)\left(c-2\right)\le 4.$
Now, if $b\ge 3$ we obtain:
$c-2\le \frac{4}{b-2}\le 4,$
which gives $c\le 6$ and the rest is smooth.
For example, the case $b=1$ is impossible, but for $b=2$ we obtain
$\frac{1}{2}=\frac{1}{a}+\frac{2}{c}$
or
$\left(b-2\right)\left(c-2\right)\le 4.$
Now, if $b\ge 3$ we obtain:
$c-2\le \frac{4}{b-2}\le 4,$
which gives $c\le 6$ and the rest is smooth.
For example, the case $b=1$ is impossible, but for $b=2$ we obtain
$\frac{1}{2}=\frac{1}{a}+\frac{2}{c}$
or
$ac=2c+4a$
or
$ac-2c-4a+8=8$
or
$\left(a-2\right)\left(c-4\right)=8$
and solve some systems:
1.$a-2=1$ and $c-4=8$ gives $\left(3,2,12\right)$
2.$a-2=2$ and $c-4=4$ gives $\left(4,2,8\right)$
3.$a-2=4$ and $c-4=2$ gives $\left(6,2,6\right)$ and
4.$a=2=8$ and $c-4=1$ gives $\left(10,2,5\right).$
Now, for $a\ge b\ge 3$$c\in \left\{3,4,5,6\right\}$, which is for you.
For the full ending of the solution it's enough to add triples $\left(b,a,c\right)$

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