Question

If xsqr+ysqr=3,ysqr +zsqr =7 and zsqr + xsqr = 19 then what is xy+zy+zx=?

Answer 1

Answer:

44.5/(sqr)^2

Step-by-step explanation:

xsqr+ysqr=3, ysqr +zsqr =7, zsqr + xsqr = 19

xy+zy+zx=?

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x+y=3/sqr, y+z=7/sqr, x+z=19/sqr

Substitute 1/sqr=m for simplicity:

1) x+y=3m, y+z=7m, x+z=19m

Subtract in pairs to get:

2) y-x=4m, z-y=12m, z-x=8m

Find x, y and z by adding pairs from 1) and 2):

y=7m/2, x=-m/2, z=31m/2

Sum of xy+zy+zx:

xy=-7m^2/4, xz=-31m^2/4, yz=216m^2/4

xy+zy+zx= m^2(216-38)/4= 44.5 m^2

Considering substitution earlier on:

44.5 m^2=44.5/(sqr)^2