What are the respective number of $\alpha$ and $\beta$- particles emitted in the following radioactive decay?

${}_{90}{}^{200}X\to {}_{80}{}^{168}Y$

Let $n-\alpha$ particles and $m-\beta$ particles are emitted. Then,

$90-2n+m=80$…(i)

$200-4n=168$        …(ii)

Solving equations (i) and (ii), we get

$n=8$ and $m=6$