Question

решите систему уравнений {(y+1)(x-12)=0 {(x-y+2)/(y-2)=5

Answer 1

(y+1)(x-12)=0
(x+y+2)/(y-2)=5

(y+1)(x-12)=0
x=y+2=5y-10

(y+1)(x-12)=0
x=6y-12

(y+1)(6y-12-12)=0

y=-1
y=4

x=6*(-1)-12
x=6*4-12

x=-18
x=12

x1=-19
y1=-1
x2=12
y2=4

(-1+1)*(-18-12)=0
-18-(-1)+2/-1-2=5

(4+1)*(12-12)=0
12-4+2/4-2=5

0=0
5=5
0=0
5=5
x1=-18
y1=-1
x2=12
y2=4