Assuming [tex]\mathrm{lg}\,x=\log_{10}x[/tex], you have from the first equation
[tex]\mathrm{lg}(x-y+1)=0\implies 10^{\mathrm{lg}(x-y+1)}=10^0[/tex]
[tex]\implies x-y+1=1\implies x-y=0\implies x=y[/tex]
From the second, you get
[tex]1+\mathrm{lg}(xy)=0\implies\mathrm{lg}(xy)=-1\implies10^{\mathrm{lg}(xy)}=10^{-1}[/tex]
[tex]\implies xy=\dfrac1{10}[/tex]
Since [tex]x=y[/tex], you have
[tex]x^2=\dfrac1{10}\implies x=y=\pm\dfrac1{\sqrt{10}}[/tex]
