you are dealing with pulleys?
can be done with addition of the two equations to eliminate T.
[tex](2^{*}) \quad {} -m_1 g \cos \theta + T = m_1 a _2[/tex]
+
[tex](4^{*}) \quad 3m_1 g - T = 3m_1 a_2 [/tex]
=
[tex](-m_1 g \cos\theta + 3m_1 g) + (T - T) = m_1 a _2 + 3m_1 a_2 \implies \\ \\
(-m_1 g \cos\theta + 3m_1 g) = 4m_1 a_2 \implies \\ \\
m_1(- g \cos\theta + 3g)= 4m_1 a_2 [/tex]
we can cancel m₁ by dividing both sides by it, assuming mass is not zero
[tex](- g \cos\theta + 3g)= 4a_2 \implies \\ \\
a_2 = \dfrac{- g \cos\theta + 3g}{4} \\ \\
a_2 = \dfrac{- 9.80 \cos 60 + 3(9.80)}{4} \\ \\
a_2 = 6.125 \text{m/s}^2
[/tex]
a₂ = 6.125 m/s² ( do significant digits if you need to)
