Simple System of Equations Part 1

It is given that:

$\begin{cases} x(y+z) = xy+zx = 39 &...(1) \\ y(z+x) = yz+xy = 60 &...(2) \\ z(x+y) = zx+yz = 63 &...(3) \end{cases}$

$\begin{aligned} Eq.3-Eq.2+Eq.1: & \quad zx+yz-yz-xy+xy+zx = 63-60+39 \\ & \quad \Rightarrow 2zx = 42 \quad \Rightarrow zx = 21 \\ Eq.1: & \quad xy + 21 = 39 \quad \Rightarrow xy = 18 \\ Eq.2: & \quad 18 + yz = 60 \quad \Rightarrow yz = 42 \end{aligned}$

$\dfrac {yz}{zx} = \dfrac {42}{21} \quad \Rightarrow \dfrac {y}{x} = 2 \quad \Rightarrow y = 2x$

$xy = 18 \quad \Rightarrow 2x^2 = 18 \quad \Rightarrow x = \pm 3 \quad \Rightarrow y = 2x = \pm 6$

$zx = 21 \quad \Rightarrow z = \pm 7$

$\Rightarrow x^2+ y^2+z^2 = 3^2+6^2+7^2 = \boxed{94}$

$\begin{cases}x(y+z)=xy+xz=39&...(1)\\y(x+z)=xy+yz=60&...(2)\\ z(x+y)= xz+yz=63 &...(3) \end{cases}$

add all 3 equation it will give $xy+yz+zx=81$ then put all 3 equation value in this new equation one by one. you will get $xy=18,yz=42,xz=21$

now multiply all these 3 new equation it will give $xyz=126$ .

now again put value of $xy,yz,xz$ in this equation you will get value of x,y and z which is 7,3,6 so value of $x^2+y^2+z^2=94$

Correct answer is: 94

x (y+z)=xy+xz(1) ,,y(x+z)=yx+yz(2),, z(x+y)=zx+zy (3),(1)+(2)=2xy+z(x+y)=99 so 2xy=99-63=36,xy =18. .,,,(2)+(3)=2zy+x(y+z)=123=2zy=123-39,zy=42,(1)+(3)=2zx+y(x+z)=102-60=42,zx =21,,so xy÷ zx =18÷21=6÷7,y =6z÷7,zy =42,so ,,z× y =z× 6z=42×7 ,,z=+or-7,y =+or-6,x=+or-3,,,,,,,,,,expression x^2 +y^2 +z^2 =94###

Adding all the first 3 equations , we get xy+xz+yz=81, so from the first, second, and third equation, we get resp. yz=42, xz=21, and xy=18. Multiply these we get xyz =+or - 126, so x=+ or -126/42, x=+ or -3, y=+ 0r -6, and z= + 0r -7. So x^2+y^2+z^2 = 94

by looking at given equations, it is evident that x,y,z has to be multiple of 3, therefore x=3, y=6, z=7.
3^2+6^2+7^2= 94