![\left \{ {{x-3y=8\, |\cdot (-2)} \atop {2x-y=6\qquad \; }} \right. \plus \left \{ {{x-3y=8} \atop {5y=-10}} \right. \; \; \left \{ {{x=3y+8} \atop {y=-2}} \right. \; \; \left \{ {{x=2} \atop {y=-2}} \right. \quad \Rightarrow \; \; (2,-2)](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7Bx-3y%3D8%5C%2C+%7C%5Ccdot+%28-2%29%7D+%5Catop+%7B2x-y%3D6%5Cqquad+%5C%3B+%7D%7D+%5Cright.+%5Cplus+++%5Cleft+%5C%7B+%7B%7Bx-3y%3D8%7D+%5Catop+%7B5y%3D-10%7D%7D+%5Cright.+%5C%3B+%5C%3B++%5Cleft+%5C%7B+%7B%7Bx%3D3y%2B8%7D+%5Catop+%7By%3D-2%7D%7D+%5Cright.+%5C%3B+%5C%3B++%5Cleft+%5C%7B+%7B%7Bx%3D2%7D+%5Catop+%7By%3D-2%7D%7D+%5Cright.+%5Cquad+%5CRightarrow+%5C%3B+%5C%3B+%282%2C-2%29)
x=2>0 , y= -2<0 ⇒ точка находится в 4 четверти
В первом билоне Х л ; во втором ( х - 5 ) л
Уравнение :
2 • ( Х - 8 ) = ( х - 5 ) + 8
2х - 16 = х + 3
Х = 19 ( л ) в первом
19 - 5 = 14 ( л ) во втором
2/5 х +12 = х/2
4х +120 = 5х
х=120
X - x2 y x y9 а другие 7б
№2
а)а(1-7)
б)4а-b²
в) 3х-2а+х=4х-2а=2(2х-а)
№3
а)6(2х-у)
б)2b(a-3c)
в)3х²(3-4у³)
№4
а)2х³-6х²у
б)6ху+4х²-9у²-6ху=4х²-9у²
в)(а²-b²)*(a+b)=а³+а²b-ab²-b³
#5
(m+2)(n-3)
х-2у-2ау+ах=х(1+а)-2у(1+а)=(х-2у)(1+а)