Ty is twice as old as dops. In 6 yrs, Ty will be 3 times what dops age was 6 yrs ago.how old are they now?

Question

Translate to a system of equations and solve. please show steps.

Answer 1

Let t = Ty's age now
Let d = Dop's age now
Then t = 2d

6 years ago, Dop was d - 6
In 6 years, Ty will be t + 6, and so
t + 6 = 3(d - 6) = 3d - 18.

Substituting the equation above for t,
2d + 6 = 3d - 18, or
6 + 18 = 3d - 2d
Thus, d = 24, and so t = 2 X 24 = 48
Ty is now 48 and Dop is now 24.

To check this, notice that 6 years ago Dop was 24 - 6 = 18.
In 6 years, Ty will be 48 + 6 = 54, and 54 = 3 X 18

Answer 2

let ty =x
let dops=y
x=2y
after 6 yrs
x=3[y-6]
x=3y-18=2y
theyfore y = 18
x=2y=2.18=36
theyfore age of ty is 36 and dops is 36

Answer 3

Age(Ty)= 2(Age(Dops))
Age(TY)+6=3((Age(Dops)-6)
Substitute The RHS of equation 1, in place of Age(Ty) in equation 2
You get:
2(Age(Dops))+6=3((Age(Dops)-6)
This can be expanded and written as:
2(Age(Dops)) +6 = 3(Age(dops))-18
Grouping the variables to the LHS and constants in the RHS we can rewrite this as:
2(Age(Dops))-3(Age(Dops))=-18-6
Simplifying:
-Age(Dops)=-24
So Age(Dops)=24
Substituting this again in equation 1,
Age(Ty)= 2(Age(Dops))= 48

Answer 4

let Ty be = "a" years old and dops be = "b" years old
2b = a...................eq 1
After 6 years: Ty = a +6
a+6 = 3b...............eq 2
Solving eq 1& 2
2b + 6 = 3b
b = 6 years
Ty is 12 years old and dops is 6 years old