Algebra Problem.HELP!! plz.?

Question

Three years ago, Mark was twice as old as Diance. Four years from now, twice Mark's age will be three less than three times Diane's age. How old are they now?
PLease solve and show me step by step.

Answer 1

---------------Mark------Diane
present---------x---------y
3 years ago--x - 3-----y - 3
4 years on-- x + 4-----y + 4

x- 3 = 2.(y - 3)
x - 3 = 2y - 6
x - 2y = - 3-------Equation 1

2.(x + 4) = 3.(y + 4) - 3
2x + 8 = 3y + 9
2x - 3y = 1------Equation 2

2x - 3y = 1
-2x + 4y = 6------ADD

y = 7
2x - 21 = 1
2x = 22
x = 11
Mark is aged 11 years.
Diane is aged 7 years.

Answer 2

let M = Mark's age,
D = Dianne's age
(M - 3) = 2(D - 3)
2(M + 4) = 3(D + 4) - 3
M - 3 = 2D - 6
2M + 8 = 3D + 9
-2M + 6 = - 4D + 12
14 = - D + 21
D = 21 - 14 = 7
M = 8 + 3 = 11

check:
2(11 + 4) = 3(7 + 4) - 3
2*15 = 3*11 - 3
30 = 33 - 3

Answer 3

Mark is 11 and Diane is 7... here's how I got that:

Let Mark be represented as m, and Diane be represented as d

From what you said, we can say that:
m-3=2(d-3)
and
2(m+4)=3(d+4)-3

If we simplify these, we get:
m+3=2d
and
2m-1=3d

We can change these to:
(m+3)/2=d
and
(2m-1)/3=d

If we set these two equations equal to each other, we get:
(m+3)/2=(2m-1)/3

This simplifies to:
m=11

We can then plug this back into one of the original equations:
11-3=2(d-3)

this simplifies to:
d=7

Answer 4

Let Mark be x
Let Diance be y
3 years ago
Y = 2X
4 years later
2X = 2Y - 3
in between the (3 years ago) and (4 years from now) is 7 years

Answer 5

M-present age of Mark, D-present age of Diance
M-3=2(D-3)
M-3=2D-6
M=2D-3 ---(1)

2*(M+4)=3*(D+4)-3
2M+8=3D+12-3
2M=3D+1 ---(2)
Plug value of M from (1) into (2)
2(2D-3)=3D+1
4D-6=3D+1
D=7

From (1)
M=2*7-3
M=11