**Question:**

Solve for x,y,z If (x+y)(y+z)=18 (y+z)(z+x)=30 (x+y)(x+z)=15

(y+z)(z+x)=30----(2)

(x+y)(x+z)=15----(15)

Now, from (1)×(2)×(3)

[(x+y)(y+z)(x+z)]^2=8100

(x+y)(y+z)(x+z)=90

Now, we have

x+y= 3---(a)

y+z=6---(b)

x+z=5----(c)

From adding (a),(b)and(c)

x+y+z=7

Hence, x=1,y=2 and =4 ans.

Give solution

X=1

Y=2

Z=4