Add the equations: $a_{n+1}+b_{n+1}=a_n+b_n$ Hence $a_n+b_n=\text{const.}=C$ Subtract the second equation from the first: $a_{n+1}-b_{n+1}=-3(a_n-b_n)$ Hence $a_n-b_n=(a_1-b_1)(-3)^{n-1}$ Thus $a_n={C+(a_1-b_1)(-3)^{n-1}\over 2}$ Assume $a_1-b_1>0$. Then observe $a_{2k}={C-(a_1-b_1)\cdot 3^{2k-1}\over 2}$. Obviously, for sufficiently large $k$ the numerator will become negative. Assume $a_1-b_1<0$. Then observe $a_{2k+1}={C+(a_1-b_1)\cdot 3^{2k}\over 2}$. Obviously, for sufficiently large $k$ the numerator will become negative. Hence $a_1-b_1=0$. QED

Adding and subtracting the two relations implies $a_n+b_n$ is constant and $a_n-b_n$ gets multiplied by $-3$ when $n$ increases by $1$. Thus if $a_1 \neq b_1$ then $a_n-b_n$ is unbounded above and below, so $a_n=\tfrac{(a_n+b_n)+(a_n-b_n)}{2}$ is unbounded above and below as well: contradiction.