Question

Four years ago Marina was three times old as
her daughter. Six years from now the mother will
be twice as old as her daughter. Find their present
ages.​

Answer 1

Answer:

here your answer..,...

Step-by-step explanation:

let Marina present age be x

and her daughter age is y

atq

4 years ago

x-4=3(y-4)

x-4=3y-12

x-3y+8=0........... (1)

after six year's

x+6=2(y+6)

x+6=2y+12

x-2y-6=0.,.,,.,...,.., (2)

equation (1)-(2)

-y+14=0

y=14

substitute

y in (1)

x-42+8=0

x-34=0

x=34

therefore

the present ages of mother and her daughter is 34,14

Answer 2

Answer:

Marina is 34 years old and her daughter is 14 years old.

Step-by-step explanation:

Gívєn -

Four years ago = Marina was three times old as her daughter.

Six years from now = the mother will

be twice as old as her daughter

Tσ fínd -

Their present Ages

Sσlutíσn -

Let the ages 4 years back be =

• Daughter as x
• Marina as 3x

$\rule{300}{1.5}$

Present Ages -

• Daughter = (x + 4)
• Marina = (3x + 4)

$\rule{300}{1.5}$

Ages after Six years from now -

• Daughter = (x + 4) + 6
• Marina = (3x + 4) + 6

$\rule{300}{1.5}$

According to the Question -

Six years from now = the mother will

be twice as old as her daughter

➡ (3x + 4) + 6 = 2(x + 4 + 6)

➡ 3x + 10 = 2x + 8 + 12

➡ 3x + 10 = 2x + 20

➡ 3x - 2x = 20 - 10

➡ x = 10

$\rule{300}{1.5}$

Present Ages:

Daughter =

➡ (x + 4)

➡ 10 + 4

➡ 14

Daughter is 14 years old.

$\rule{300}{1.5}$

Marina =

➡ 3x + 4

➡ 3(10) + 4

➡ 30 + 4

➡ 34

Marina is 34 years old.

$\therefore$ Marina is 34 years old and her daughter is 14 years old.