Question

Help sole these equations 1) 3y + 1 =16 ???? 2) 18=4k+6 ????

Answer 1

$1)\\ 3y+1=16\ \ \ \ | subtract\ 1\\\\ 3y=15\ \ \ \ | divide\ by\ 3\\\\ y=5\\\\\ 2)\\ 18=4k+6\ \ \ \ | subtract\ 6\\\\ 12=4k\ \ \ \ | divide\ by\ 4\\\\ k=3$

Answer 2

3y+1=16
Subtract 1 from both sides
(3y+1)-1=(16)-1
3y=15
Divide both sides by 3
(3y)/3=(15)/3
Y=5

18=4k+6
Subtract 6 from both sides
(18)-6=(4k+6)-6
12=4k
Divide both sides by 4
(12)/4=(4k)/4
3=k

Final answer:
1) y=5
2) k=3