Question 974586: Solve for x,y,z if
(x+y)(y+z)=18,(y+z)(z+x)=30and(x+y)(x+z)=15
Answer by amarjeeth123(569)   (Show Source):
You can put this solution on YOUR website! (x+y)(y+z)=18........equation 1
(y+z)(z+x)=30........equation 2
(x+y)(x+z)=15........equation 3
Multiplying the three equations we get,
[(x+y)(y+z)(z+x)]^2=18(30)(15)=8100
(x+y)(y+z)(z+x)=√8100=90....equation 4
Dividing equation 4 by equation 1 we get
z+x=5......equation 5
Dividing equation 4 by equation 2 we get
x+y=3......equation 6
Dividing equation 4 by equation 3 we get
y+z=6......equation 7
Adding equations 5,6 and 7 we get,
2(x+y+z)=14
x+y+z=7....equation 8
Subtracting equation 5 from equation 8
y=2
x=1
z=4
x=1,y=2 and z=4
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