Two years ago a woman wad 7 times as old as her daughter, but in 3 years time she would be x times as old as the girl. how old are they now?

The present age of the woman is 37 years and her daughter is 7 years

Step-by-step explanation:

two years ago a woman was 7 times old as her daughter, but in 3 years time she would be 4 times old as the girl. how old are they now

Two years ago a woman wad 7 times as old as her daughter

Let her daughter=x-2

The woman=y-2

x-2

7(y-2)=7y-14

x-2=7y-14

x-7y=-14+2

x-7y= -12 (1)

but in 3 years time she would be 4 times as old as the girl.

x+3

y+3

x+3

4(y+3)=4y+12

x+3=4y+12

x-4y=12-3

x-4y=9. (2)

x-7y= -12 (1)

x-4y=9. (2)

Subtract (1) from (2)

4y-(-7y)=9-(-12)

-4y+7y=9+12

3y=21

y=21/3

y=7

Substitute

y=7 into (1)

x-7y= -12

x-7(7)=-12

x-49=-12

x= -12+49

=37

The present age of the woman is 37 years and her daughter is 7 years

3: 2

step-by-step explanation:

shorter piece is 2: 3, so just reverse it, and you get 3: 2

cosine

step-by-step explanation: