System of equations (contest problem)

Question

Compute the ordered triple $(x,y,z)$ of positive real numbers that satisfies all three of the equations:

$xy+x+y=19$

$yz+y+z=29$

$xz+x+z=53$

Please show me specific work and explain the law or principle used. Thank you so much!!!

Answer 1

By adding $1$ on both sides of equations we get ( That's just a small trick for making factorization easy) $$xy+x+y+1=20=(x+1)(y+1)$$ $$yz+y+z+1=30=(z+1)(y+1)$$ $$xz+x+z+1=54=(x+1)(z+1)$$

Now put $$A=x+1$$$$B=y+1$$$$C=z+1$$ $$AB=20$$ $$BC=30$$ $$AC=54$$

By solving using usual methods we get

$$x=5,y=\frac{7}{3},z=8$$

So the required triplet is $$\left(5,\frac{7}{3},8\right)$$

Answer 2

$xy+x+y+1=20$
$yz+y+z+1=30$
$xz+x+z+1=54$
$(x+1)(y+1)=20$ (1)
$(y+1)(z+1)=30$ (2)
$(x+1)(z+1)=54$ (3)
$(1)/(2)$:
$$(x+1)/(z+1)=2/3$$ $$(x+1)=2/3*(z+1)$$ $$(3):2/3*(z+1)^2=54$$ $z=8;$then $x=5;y=7/3$
$z=-10;$then $x=-7;y=-13/3$ but they are not positive . So answer is only $(5;7/3;8)$

Answer 3

If you subtract the first equation from the third one you will have $(x+1)(z-y)=34$. In the similar manner if you subtract the second one from the third, then you have $(z+1)(x-y)=24$, and from subtraction of the first one from the second one we have $(y+1)(z-x)=10$, hence till now:

$(x+1)(z-y)=34\space \space (1)$

$(z+1)(x-y)=24\space \space (2)$

$(y+1)(z-x)=10\space \space (3)$

Now by multiplication of the first, second, and the third equations by $z,x$ and $y$ respectively we have:

$xyz+xz+yz=19z$

$xyz+xz+yx=29x$

$xyz+xy+yz=53y$

By subtracting the first above equation from the third one, we have:

$x(y-z)=53y-19z\space \space (4)$

By subtracting the second above equation from the third one, we have:

$z(y-x)=53y-29x\space \space (5)$

By subtracting the first above equation from the second one, we have:

$y(x-z)=29x-19z\space \space (6)$

Now by substituting $4$ in $1$ we have:

$20z-54y=34\space \space (7)$

In the similar manner we have:

$30x-54y=24\space \space (8)$

$20z-30x=10\space \space (9)$

Note that the equations $7,8$ and $9$ are not independent, so we cannot obtain parameters from them, but by substituting corresponding values for $y$ and $z$, in terms of $x$, obtained from $8$ and $9$ we have:

$z=\frac{3x+1}{2}$ and $y=\frac{15x-12}{27}$.

By substituting these values in the first equation in question we reach to the equation $x^2+2x-35=0$, which have the solutions $x=-7$ and $x=5$, hence there are two sets of triples as below:

$(5,\frac{7}{3},8)$ and $(-7,\frac{-13}{3},-10)$