If ₉₂U²³⁵ assumed to decay only by emitting two $\alpha$- and one $\beta$-particles, the possible product of decay is 

${ }_{92}{\mathrm{U}}^{235}{⟶}_Z{\mathrm{X}}^A+2_2{\mathrm{He}}^4{+}_{-1}{e}^0$

Equating mass number of both sides

235 = A + 2 × 4 + 0
$\therefore$A = 235 - 8 = 227

Similarly, equating atomic number of both sides

$\begin{array}{l}92=Z+2\times 2-1\\ \therefore Z=92-3=89\\ \therefore Z^A{=}_{89}{\mathrm{Ac}}^{227}\end{array}$