While solving some temperature questions as to physics, I had found an equation which I couldn't solve for $T$, where $T$ is the final temperature of water.
$$3m_1c_1(T-30)+m_2c_2(T-10)=0$$
If $m_1c_1 = m_2c_2$, how to solve it for the unknown $T$?
Let me show my attempt
$$3(T-30) + (T-10) = 0\implies3T - 90 +T -10 = 0$$
$$4T = 100 \implies T = \boxed{25}$$
You are perfectly correct.
$3m_1c_1(T-30)+m_2c_2(T-10)=0$
$\implies T(3m_1c_1+m_2c_2)+(-90m_1c_1-10m_2c_2) =0$
$\implies T(4m_1c_1) +(-100m_1c_1) = 0$
$\implies T =\dfrac{100m_1c_1}{4m_1c_1}$
$\implies T = 25$