Question

已知y=2y1+y2,y1与x成正比例,y2与x+2成正比例,且当x=1,y=3;x=-3时,y=-21.求y余x之间函

Answer 1

已知y=2y1+y2,y1与x成正比例,y2与x+2成正比例,且当x=1,y=3;x=-3时,y=-21

y1=k1x

y2=k2(x+2)

y=2k1x+k2(x+2)

当x=1,y=3;

3=2k1*1+k2(1+2)

2k1+3k2=3 (1)

x=-3时,y=-21

-21=2k1*(-3)+k2(-3+2)

6k1+k2=21 (2)

(2)*3-(1)

18k1+3k2-2k1-3k2=3*21-3

16k1=60

k1=15/4

k2=-3/2

y=2*(15/4)x-3/2(x+2)

y=15/2x-3/2x-3

y=6x-3