8х-2у=у^2, если у=6х-1

[latex]\begin{cases}y=6x-1\\ 8x-2y=y^2\end{cases}[/latex] [latex]8x-2(6x-1)=(6x-1)^2 \\ 8x-12x+2=36x^2-12x+1 \\ -4x+2=36x^2-12x+1 \\ 36x^2-8x-1=0 \\ D=64-4*36*(-1)=64+144=208 \\ x= \frac{8б \sqrt{208} }{72} = \frac{8б \sqrt{16*13} }{72} = \frac{8б4 \sqrt{13} }{72} = \frac{4(2б \sqrt{13} )}{72} = \frac{2б \sqrt{13} }{18}[/latex] Ответ:  [latex]\frac{2б \sqrt{13} }{18}[/latex]