[tex]42_x+53_x=125_x\\4\cdot x^1+2\cdot x^0+5\cdot x^1+3\cdot x^0=1\cdot x^2+2\cdot x^1+5\cdot x^0\\4x+2+5x+3=x^2+2x+5\\x^2-7x=0\\x(x-7)=0\\x=0 \vee x=7[/tex]
There is no numeral system with base 0, so [tex]x=7[/tex].
