A nice algebra question | System of equations with three variables  

Literally as easy as 123...

x(y+z)=5……(1) y(x+z)=8……(2) z(x+y)=9……(3) (1)+(2)+(3)⇒2(xy+yz+zx)=22 ⇒xy+yz+zx=11……(4) (4)-(1)⇒yz=6……(5) (4)-(2)⇒zx=3……(6) (4)-(3)⇒xy=2……(7) (5)×(6)×(7)⇒(xyz)^2=36 ⇒xyz=±6……(8) (8)÷(5)⇒x=±1 (8)÷(6)⇒y=±2 (8)÷(7)⇒z=±3

Wayyyy too long....!

Sorry gentleman. Your efforts are good. But not teaching a technique. Whenever you have a+b, b+c, c+a or ab, bc, ca, for the first one take the sum and subtract each of a+b, b+c, c+a one by one. You will get a,b,c. Similarly for the second one multiply ab, bc, ca and take square root. Then divide by each one you will get a,b,c. In the present problem, use both the techniques. You will get the answers just like that