Problem: If 2x−3y−z=02 x-3 y-z=02x−3y−z=0 and x+3y−14z=0,z≠0x+3 y-14 z=0, z \neq 0x+3y−14z=0,zî€ =0, the numerical value of
x2+3xyy2+z2\dfrac{x^{2}+3 x y}{y^{2}+z^{2}} y2+z2x2+3xy​
is:
Answer Choices:
A. 777
B. 222
C. 000
D. −20/17-20 / 17−20/17
E. −2-2−2 Solution:
2x−3y=22 x-3 y=22x−3y=2
x+3y=14z∴x=5z,y=3z∴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=14z∴x=5z,y=3z∴y2+z2x2+3xy​=10z270z2​=7