Answer:
D (12, 21); B (11, 17); C (18,28); D (42); C (8). 
Step-by-step explanation:
First: 
Let I represent Isabel and M represent Marie. 
We know that currently, Isabel is 9 years older than Marie, or: 
I=9+M. 
In three years, Isabel will be six years less than twice of Marie's age. In other words: 
(I+3)= 2(M+3)-6
Now solve. Substitute I. 
(9+M+3)=2M+6-6
12+M=2M
M=12; I=21. Marie is 12 while Isabel is 21. 
Second:
Let I represent Isabel and M represent Marie. 
Isabel is 6 years older than Marie; in other words: I=6+M
In 4 years, Isabel will be 9 years less than twice Marie's age. Or: 
(I+4)=2(M+4)-9
Solve. Substitute I. 
(6+M+4)=2(M+4)-9
10+M=2M+8-9
10+M=2M-1
11=M; I=17; Marie is 11 while Isabel is 17. 
Third: 
Let I represent Isabel and M represent Marie. 
Isabel is 10 years older than Marie, or: I=10+M
In 2 years, twice Isabel's age is three times Marie's age. Or: 
2(I+2)=3(M+2)
Solve. Substitute for I. 
2(10+M+2)=3(M+2)
24+2M=3M+6
18=M; I=28. Marie is 18 while Isabel is 28. 
Fourth: 
Let M represent Mary and A represent Ann. 
Mary is 3 time as old as Ann. Or: M=3A
7 years ago Mary was 5 times as old as Ann. In other words: 
(M-7)=5(A-7)
Solve for this system. Substitute M. 
(3A-7)=5A-35
-2A=-28
A=14; M=42; Mary is 42. 
Fifth: 
Let T represent Tammy and L represent Laurel. 
We know that Tammy is 42 while Laurel is 9. In other words: 
T=42 and L=9. 
We need to find in how many years will 3 times Laurel's age be 1 more than Tammy's age. In other words, let's let y represent the amount of years. Thus: 
3(L+y)=(T+y)+1
We already know L and T: 
3(9+y)=(42+y)+1
27+3y=43+y
2y=16
y=8
In 8 years, when Laurel is 17 and Tammy is 50. (3 times 17 is 51, one more than 50).