In the chemical equation MnO₂ + xHCl → MnCl₂ + yH₂O + zCl₂, what are the values of x, y, and z?

What you are solving here

You are finding the coefficients $x$, $y$, and $z$ that balance the number of atoms of each element on both sides of the reaction.

Balance oxygen using water

There are 2 oxygen atoms in $\mathrm{MnO_2}$, so you need 2 oxygens on the products side. Put 2 in front of water:

$$\mathrm{MnO_2 + xHCl \rightarrow MnCl_2 + 2H_2O + zCl_2}$$

Now oxygen is balanced.

Balance hydrogen using HCl

With $\mathrm{2H_2O}$ you have $4$ hydrogen atoms on the right, so you need $4$ hydrogens on the left. Since each $\mathrm{HCl}$ gives 1 H, set $x=4$:

$$\mathrm{MnO_2 + 4HCl \rightarrow MnCl_2 + 2H_2O + zCl_2}$$

Balance chlorine and solve for z

Left side chlorine atoms: $4$ (from $\mathrm{4HCl}$).

Right side chlorine atoms come from $\mathrm{MnCl_2}$ (which has 2 Cl) plus $\mathrm{zCl_2}$ (which has $2z$ Cl):

$$2 + 2z = 4 \;\Rightarrow\; 2z = 2 \;\Rightarrow\; z = 1$$

Final balanced equation

$$\mathrm{MnO_2 + 4HCl \rightarrow MnCl_2 + 2H_2O + Cl_2}$$

So $x=4$, $y=2$, $z=1$.