3x2−5y3=−2 ...(i)x3+y2=136 ...(ii)From equation (i), we get,9x−10y=−12x=−129+10y9 ...(iii)Putting this value in equation (ii), we get,
−12+10y93+y2=136−12+10y27+y2=136−24+20y+27y54=13647y=117+2447y=141y=3 ...(iv)Substituting this value in equation (iii), we obtain,x=−12+10×39=189=2Hence,x=2,y=3