coccassi function algebra

In the first equation, we divide both sides by $xy$ . Yielding

$\dfrac{1}{x}+\dfrac{1}{y}=5$ .

Doing the same thing to the other equations give:

$\dfrac{1}{y}+\dfrac{1}{z}=6$ and

$\dfrac{1}{z}+\dfrac{1}{x}=7$

For equations 2 and 3, respectively.

Adding our new equations, we get

$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=9$

Subtracting $\dfrac{1}{x}+\dfrac{1}{y}=5$ from the new equation yields

$\dfrac{1}{z}=4$ .

Doing the same thing to $x$ and $y$ , we get:

$\dfrac{1}{x}=3$ , and

$\dfrac{1}{y}=2$ .

Now we know that $x=\dfrac{1}{3}$ , $y=\dfrac{1}{2}$ , and $z=\dfrac{1}{4}$ ,

$x+y+z=\dfrac{13}{12}$ ,

And $13+12=\boxed{25}$

$x+y=5xy\\ solve\quad x\quad in\quad terms\quad of\quad y\\ \frac { x+y }{ xy } =5\\ \frac { 1 }{ y } +\frac { 1 }{ x } =5\\ \frac { 1 }{ x } =5-\frac { 1 }{ y } \\ x=\frac { y }{ 5y-1 } \longrightarrow \quad \left( 1 \right) \\ \\ do\quad the\quad same\quad at\quad y+z=6yz\\ solve\quad z\quad in\quad terms\quad of\quad y,\quad we\quad can\quad get\\ z=\frac { y }{ 6y-1 } \longrightarrow \quad \left( 2 \right) \\ \\ then\quad substitute\quad (1)\quad and\quad (2)\quad to\quad the\quad 3rd\quad given\quad equation\\ z+x=7zx\\ \frac { y }{ 6y-1 } +\frac { y }{ 5y-1 } =7\left( \frac { y }{ 6y-1 } \right) \left( \frac { y }{ 5y-1 } \right) \\ \\ y=0.5\\ \\ then\quad substitute\quad the\quad value\quad of\quad y\quad to\quad equation\quad (1)\quad and\quad (2)\quad to\quad get\quad the\quad value\quad of\quad x\quad and\quad z\\ x=\frac { 1 }{ 3 } \quad and\quad z=\frac { 1 }{ 4 } \quad \\ then\quad add\quad x,y,z\\ x+y+z\quad =\quad \frac { 13 }{ 12 } \\ therefore\quad the\quad answer\quad will\quad be\quad 13+12\quad =\quad 25$

Multiply first equation by $x$ :

$zx+zy=5xyz$

Similarly:

$xy+xz=6xyz$

$yz+yx=7xyz$

Summing the three equations we get:

$xz+zy+xy=9xyz$

Subtracting equation 1 we obtain $xy=4xyz$ , then $z=\frac{1}{4}$

Similarly, by subtracting equations 2 and 3 from equation 4, respectively, we get $x=\frac{1}{3}$ and $y=\frac{1}{2}$ .

Summing up, $x+y+z=\frac{13}{12}$ , so the answer is 25

x+y=5xy...divide both sides by xy giving 1/x + 1/y = 5...similarly 1/y + 1/z =6 & 1/x +1/z = 7...now let 1/x=a,1/y=b & 1/z =c...so the new equations are a+b=5 , b+c=6 & c+a=7...three equations and three unknowns solve it.finally a=3 gives 1/x = 3 so x=1/3.similarly y =1/2 & z=1/4

There are 3 equation and 3 variable
Minus first and second equation put the relationship in first

Then question will be solved If you have problem please comment