Решите систему уравнений!x/y+y/x=17/4x+y=10

[latex] \frac{x}{y} =t \\ \frac{y}{x}= \frac{1}{t} \\ t+ \frac{1}{t} = \frac{17}{4} \\ \frac{t^2+1}t} =\frac{17}{4} \\ \frac{t^2+1}t}-\frac{17}{4}=0 \\ \frac{4t^2+4-17t}{4t} =0 \\t \neq 0 \\ 4t^2+4-17t=0 \\ D=(-17)^2-4*4*4=289-64=225=15^2 \\ t1= \frac{17+15}{2*4} =4 \\ t2= \frac{17-15}{2*4} = \frac{1}{4} \\ \frac{x1}{y1} =4 \\ \frac{x2}{y2} =0.25 \\ x1=4y1 \\ 4y1+y1=10 \\ 5y1=10 \\ y1=2 \\ x1=4*2=8 \\ x2=0.25y2 \\ 0.25y2+y2=10 \\ 1.25y2=10 \\ y2=8 \\ x2=0.25*8=2[/latex]