Ответ:
координаты точки пересечения (2;-3)
Объяснение:
Можно решить как систему линейных уравнений:
![\left \{ {{3x-2y=12} \atop {5x+3y=1}} \right. \\\\\\\left \{ {{9x-6y=36} \atop {10x+6y=2}} \right. \\\\19x=38\\\\x=38:19\\\\x=2 \left \{ {{3x-2y=12} \atop {5x+3y=1}} \right. \\\\\\\left \{ {{9x-6y=36} \atop {10x+6y=2}} \right. \\\\19x=38\\\\x=38:19\\\\x=2](https://tex.z-dn.net/?f=%5Cleft%20%5C%7B%20%7B%7B3x-2y%3D12%7D%20%5Catop%20%7B5x%2B3y%3D1%7D%7D%20%5Cright.%20%20%20%5C%5C%5C%5C%5C%5C%5Cleft%20%5C%7B%20%7B%7B9x-6y%3D36%7D%20%5Catop%20%7B10x%2B6y%3D2%7D%7D%20%5Cright.%20%5C%5C%5C%5C19x%3D38%5C%5C%5C%5Cx%3D38%3A19%5C%5C%5C%5Cx%3D2)
5*2+3y=1
3y=1-10
3y= -9
y= -9:3
y= -3
(2;-3)
Можно выразить у через х в обеих функциях и,приравняв их,найти координаты точки пересечения:
2y=3x-12
y=(3x-12)/2
3y=1-5x
y=(1-5x)/3
![\frac{3x-12}{2} =\frac{1-5x}{3} \\\\3(3x-12)=2(1-5x)\\\\9x-36=2-10x\\\\9x+10x=2+36\\\\19x=38\\\\x=38:19\\\\x=2 \frac{3x-12}{2} =\frac{1-5x}{3} \\\\3(3x-12)=2(1-5x)\\\\9x-36=2-10x\\\\9x+10x=2+36\\\\19x=38\\\\x=38:19\\\\x=2](https://tex.z-dn.net/?f=%5Cfrac%7B3x-12%7D%7B2%7D%20%3D%5Cfrac%7B1-5x%7D%7B3%7D%20%5C%5C%5C%5C3%283x-12%29%3D2%281-5x%29%5C%5C%5C%5C9x-36%3D2-10x%5C%5C%5C%5C9x%2B10x%3D2%2B36%5C%5C%5C%5C19x%3D38%5C%5C%5C%5Cx%3D38%3A19%5C%5C%5C%5Cx%3D2)
y₁= ![\frac{3x-12}{2} =\frac{3*2-12}{2} =\frac{-6}{2} =-3 \frac{3x-12}{2} =\frac{3*2-12}{2} =\frac{-6}{2} =-3](https://tex.z-dn.net/?f=%5Cfrac%7B3x-12%7D%7B2%7D%20%3D%5Cfrac%7B3%2A2-12%7D%7B2%7D%20%3D%5Cfrac%7B-6%7D%7B2%7D%20%3D-3)
y₂= ![\frac{1-5x}{3} =\frac{1-5*2}{3} =\frac{-9}{3} = -3 \frac{1-5x}{3} =\frac{1-5*2}{3} =\frac{-9}{3} = -3](https://tex.z-dn.net/?f=%5Cfrac%7B1-5x%7D%7B3%7D%20%3D%5Cfrac%7B1-5%2A2%7D%7B3%7D%20%3D%5Cfrac%7B-9%7D%7B3%7D%20%3D%20-3)