Чтобы найти точки пересечения, нужно объединить два уравнения в систему.
![\left \{ {{x^2+y^2=17} \atop {5x=17+3y}} \right. \\ \left \{ {{( \frac{17+3y}{5} )^2+y^2=17} \atop {x=\frac{17+3y}{5}}} \right. \\ \left \{ {{ \frac{289+102y+9y^2}{25}+y^2-17=0 } \atop {x=\frac{17+3y}{5}}} \right. \\ \left \{ {{289+102y+9y^2+25y^2-425=0} \atop {x=\frac{17+3y}{5}}} \right. \\ \left \{ {{34y^2+102y-136=0} \atop {x=\frac{17+3y}{5}}} \right. \\ \left \{ {{y^2+3y-4=0} \atop {x=\frac{17+3y}{5}}} \right. \\ \left \{ {{y=-4} \atop {x=1}} \right. \left \{ {{y=1} \atop {x=4}} \right.](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7Bx%5E2%2By%5E2%3D17%7D+%5Catop+%7B5x%3D17%2B3y%7D%7D+%5Cright.++%5C%5C++%5Cleft+%5C%7B+%7B%7B%28+%5Cfrac%7B17%2B3y%7D%7B5%7D+%29%5E2%2By%5E2%3D17%7D+%5Catop+%7Bx%3D%5Cfrac%7B17%2B3y%7D%7B5%7D%7D%7D+%5Cright.+%5C%5C++%5Cleft+%5C%7B+%7B%7B+%5Cfrac%7B289%2B102y%2B9y%5E2%7D%7B25%7D%2By%5E2-17%3D0+%7D+%5Catop+%7Bx%3D%5Cfrac%7B17%2B3y%7D%7B5%7D%7D%7D+%5Cright.+%5C%5C+%5Cleft+%5C%7B+%7B%7B289%2B102y%2B9y%5E2%2B25y%5E2-425%3D0%7D+%5Catop+%7Bx%3D%5Cfrac%7B17%2B3y%7D%7B5%7D%7D%7D+%5Cright.++%5C%5C++%5Cleft+%5C%7B+%7B%7B34y%5E2%2B102y-136%3D0%7D+%5Catop+%7Bx%3D%5Cfrac%7B17%2B3y%7D%7B5%7D%7D%7D+%5Cright.++%5C%5C++%5Cleft+%5C%7B+%7B%7By%5E2%2B3y-4%3D0%7D+%5Catop+%7Bx%3D%5Cfrac%7B17%2B3y%7D%7B5%7D%7D%7D+%5Cright.++%5C%5C++%5Cleft+%5C%7B+%7B%7By%3D-4%7D+%5Catop+%7Bx%3D1%7D%7D+%5Cright.++%5Cleft+%5C%7B+%7B%7By%3D1%7D+%5Catop+%7Bx%3D4%7D%7D+%5Cright.+)
Ответ: (1; -4), (4; 1)
1) 17+9+8=35 (игр.) всего
2) 35-25=10 (игр.)
Ответ:10 игрушек отнесли в 2 группу
X+5X=60
<span>6X=60 </span>
<span>x=10 </span>
<span>один 10, другой 5*10=50</span>
=(1/5)^20· 5^21 + (2·5)^6 :10^5=5^ (-20)· 5^21 +10^6:10^5= 5+10=15.