Answers:
1) First table: y=2x+24
2) Second table: y=(3/2)x+24
Solution:
m=(y2-y1)/(x2-x1)
y-y1=m(x-x1)
1) First table. Taking the first two points of the table:
P1=(-6,12)=(x1,y1)→x1=-6, y1=12
P2=(-5,14)=(x2,y2)→x2=-5, y2=14
m=(y2-y1)/(x2-x1)
m=(14-12)/(-5-(-6))
m=(2)/(-5+6)
m=(2)/(1)
m=2
y-y1=m(x-x1)
y-12=2(x-(-6))
y-12=2(x+6)
y-12=2x+2(6)
y-12=2x+12
y-12+12=2x+12+12
y=2x+24
2) Second table. Taking the first two points of the table:
P1=(-14,3)=(x1,y1)→x1=-14, y1=3
P2=(-12,6)=(x2,y2)→x2=-12, y2=6
m=(y2-y1)/(x2-x1)
m=(6-3)/(-12-(-14))
m=(3)/(-12+14)
m=(3)/(2)
m=3/2
y-y1=m(x-x1)
y-3=(3/2)(x-(-14))
y-3=(3/2)(x+14)
y-3=(3/2)x+(3/2)(14)
y-3=(3/2)x+(3)(14)/2
y-3=(3/2)x+42/2
y-3=(3/2)x+21
y-3+3=(3/2)x+21+3
y=(3/2)x+24