We have x+y-7=0 x + y = 7 ...(i) and 3x + y - 13 = 0 = 3x + y = 13 ...(ii) B subtracting Eq. (i)from Eq. (ii), we get 3x + y = 13 ...(i) x + y = 7 ...(iI) 2x = 6 Therefore, x = 3 On putting the value of x in q. (i), we get 3 + y = 7 Therefore, y = 4 Now, \( \Large 4x^{2}+y^{2}+4xy \) \( \Large 4 \times \left(3\right)^{2}+ \left(4\right)^{2}+4 \times 3 \times 4 \) \( \Large 4 \times 9+16+48 \) \( \Large 36+16+48=100 \)