Can you solve these equations?

$\begin{cases} xy+yz = 35 &...(1) \\ yz+zx = 32 &...(2) \\ zx+xy = 27 &...(3) \end{cases}$

$\Rightarrow \begin{cases} (1)+(2)+(3): & 2(xy+yz+zx) = 94 \\ & \Rightarrow xy+yz+zx = 47 & ... (4) \\ (4)-(1): & zx = 47-35 = 12 & ...(5) \\ (4)-(2): & xy = 47-32 = 15 & ...(6) \\ (4)-(3): & yz = 47-27 = 20 & ...(7) \end{cases}$

$\Rightarrow \begin{cases} \dfrac{(6)\times (5)}{(7)}: & \dfrac {xy \times zx} {yz} = \dfrac {15\times 12} {20} & \Rightarrow x^2 = 9 & \Rightarrow x = \pm 3 \\ (6): & \pm 3 y = 15 & \Rightarrow y = \pm 5 \\ (5): & \pm 3 z = 12 & \Rightarrow z = \pm 4 \end{cases}$

$\Rightarrow x^2 + y^2 + z^2 = 9 + 25 + 16 = \boxed{50}$

First time writing a solution.. Hope this helps

Let (i) xy + yz = 35 (ii) yz + xz = 32 (iii) xz + yx = 27

Calculate (i) - (ii) + (iii) = (xy - xz) + (xz + yx) = 2xy = 30 xy = 15

(ii) - (iii) + (i) = (yz - yx) + (xy + yz) = 2yz = 40 yz = 20

(iii) - (i) + (ii) = (xz - yz) + (yz + xz) = 2xz = 24 xz = 12

By inserting x = 15/y and y = 20/z, you get z = 4

Then xz = 12 so x = 3

And xy = 15 so y = 5

Therefore, x^2 + y^2 + z^2 = 9 + 25 + 16

= 50

The sum of $35+32+27=94$ , which is $2(xy+yz+xz)$ .

$xy+yz+xz=\frac{94}{2} = 47$

$xz$ = $47-35=12$

$xy$ = $47-32=15$

$yz$ = $47-27=20$

$HCF(12,15) = 3$

$HCF(12,20) = 4$

$HCF(20,15) = 5$

$3^{2} + 4^{2} + 5^{2} = \boxed{50}$

xy+ yz= 35...(1)
yz+ xz= 32...(2)
xz+ yx= 27...(3)
((1)+ (2)+ (3))/ 2...xy+ yz+ zx= 47..(4)
(4)- (1)...zx= 12...(5)
(4)- (2)...yx= 15...(6)
(4)- (3)...yz= 20...(7)
(5)* (6)* (7)...(xyz)^2= 12* 15* 20...or (xyz)= 3 4 5...(8)
(8)/ (5)...y= 5
(8)/ (6)...z= 4
(8)/ (7...x= 3
so...x^2+ y^2+ z^2= 9+ 25+ 16= 50