Problem: If 2xβ3yβz=02 x-3 y-z=02xβ3yβz=0 and x+3yβ14z=0,zβ 0x+3 y-14 z=0, z \neq 0x+3yβ14z=0,zξ =0, the numerical value of
x2+3xyy2+z2\dfrac{x^{2}+3 x y}{y^{2}+z^{2}} y2+z2x2+3xyβ
is:
Answer Choices:
A. 777
B. 222
C. 000
D. β20/17-20 / 17β20/17
E. β2-2β2 Solution:
2xβ3y=22 x-3 y=22xβ3y=2
x+3y=14zβ΄x=5z,y=3zβ΄x2+3xyy2+z2=70z210z2=7x+3 y=14 z \quad \therefore x=5 z, y=3 z \quad \therefore \dfrac{x^{2}+3 x y}{y^{2}+z^{2}}=\dfrac{70 z^{2}}{10 z^{2}}=7x+3y=14zβ΄x=5z,y=3zβ΄y2+z2x2+3xyβ=10z270z2β=7