Consider the following reactions Na_ 2 ~B_ 4 O_ 7 Δ 2 X+ Y CuSO_ 4...

The given reactions describe the Borax Bead Test.
When borax ($Na_2B_4O_7 \cdot 10H_2O$) is heated, it first loses water and then melts to form a transparent glassy bead consisting of sodium metaborate ($NaBO_2$) and boric anhydride ($B_2O_3$).
$Na_2B_4O_7 \xrightarrow{\Delta} 2NaBO_2 (X) + B_2O_3 (Y)$
In the non-luminous (oxidizing) flame, $B_2O_3$ reacts with $CuSO_4$ to form copper(II) metaborate ($Z$), which is blue in color.
$CuSO_4 + B_2O_3 \xrightarrow{Non-Luminous} Cu(BO_2)_2 (Z) + SO_3$
In $Cu(BO_2)_2$, the oxidation state of $Cu$ is $+2$.
In the luminous (reducing) flame, the copper(II) metaborate is reduced by carbon to copper(I) metaborate ($Q$), which is colorless/red.
$2Cu(BO_2)_2 (Z) + 2NaBO_2 (X) + C \xrightarrow{Luminous} 2CuBO_2 (Q) + Na_2B_4O_7 + CO$
In $CuBO_2$, the oxidation state of $Cu$ is $+1$.
Therefore, the oxidation states of $Cu$ in $Z$ and $Q$ are $+2$ and $+1$ respectively.