A mother is twice as old as her daughter now, but ten years ago, she was 3 times as old as her daughter. Fins their ages at present.
I tried to form an equation: 2x-10=3x, but that didn't work. Could someone help me forma working equation for it, and I can take it from there.
Thank you!
Michelle
M=2D
M-10=3(D-10)
the second one becomes
M=3(D-10)+10
set this equation equal to the first, and solve for D.
That does not help!
To find their ages at present, we can set up two equations based on the given information.
Let's denote the mother's age as "M" and the daughter's age as "D."
1. The problem states that the mother is currently twice as old as her daughter, so we can write the equation: M = 2D.
2. It is also stated that ten years ago, the mother was three times as old as her daughter. So, we can set up the equation: M - 10 = 3(D - 10).
Now, we can solve the system of equations to find the values of M and D.
Substitute the value of M from the first equation into the second equation:
2D - 10 = 3(D - 10).
Expand and simplify:
2D - 10 = 3D - 30.
Rearrange the equation by bringing all the D terms to one side:
2D - 3D = -30 + 10.
Combine like terms:
-D = -20.
Multiply both sides by -1 to isolate D:
D = 20.
Now, substitute the value of D into the first equation to find M:
M = 2(20).
M = 40.
Therefore, the daughter's age is 20 and the mother's age is 40 at present.