Solve the following systems of equations:
Equating both equations
⇒ 6u – 2v = 2u + 6v
⇒ u = 2v
Substituting value of u
⇒ 2(6v – v) = 5 × 2v × v
⇒ v = 1
Thus, 2(3u – 1) = 5u
⇒ 6u – 2 = 5u
⇒ u = 2