The correct option is B 7,9
Given,
x+y2+3x−5y4=2.............Eq(1)
x14+y18=1....................Eq(2)
On solving eq (1) we get
x+y2+3x−5y4=2
⇒2(x+y)+(3x−5y)4=2
⇒2x+2y+3x−5y=8
⇒5x−3y=8.............................Eq(3)..multiplied by 7
on solving Eq(2) we get
x14+y18=1
⇒9x+7y126=1
⇒9x+7y=126..........................Eq(4)...multipied by3
on adding both Eq 3 and Eq 4 we get
35x−21y=56
27x+21y=378
_________________________
62x=434
x=43462=7
putting the value of x=7 in eq 4 we get
9×7+7y=126
7y=126−63
7y=63
y=637=9
Hence x=7 and y=9