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).
