Question

(1) Two years ago, the age of a father was three and a half times the age of his
daughter then. Six years hence, the age of father will be ten years more than twice
the age of his daughter then. Find their present ages.​

Answer 1

father's age = x = 44 years

daughters age = y = 14 years

Step-by-step explanation:

let father's age be = x

daughters age be = y

two years ago father's age = x - 2

two years ago daughters age = y - 2

father was 3 1/2 times daughter age

6 years later

father age = x + 6

daughters age = y + 6

father will be 10 years more than twice of daughters age

\begin{gathered}x + 6 = 2(y + 6) + 10 \\ x + 6 = 2y + 12 + 10 \\ x + 6 = 2y + 22 \\ x - 2y - 16 = 0 \\ x - 2y = 16\: \: \: \: equ \: (2)\end{gathered}

x+6=2(y+6)+10

x+6=2y+12+10

x+6=2y+22

x−2y−16=0

x−2y=16equ(2)

multiply equ (2) with 2

2x - 4y = 32. equ (3)

subtract (3) - (1)

2x - 7y = -10

-(2x - 4y = 32)

we get

-3y = - 42

y = 42/3

y = 14

substitute y = 14 in equ (2)

x - 2y = 16

x - 2(14) = 16

x - 28 = 16

x = 16 + 28

x = 44

therefore father's age = x = 44 years

daughters age = y = 14 years

hope you get your answer

Answer 2

Answer:

14 and 44

Step-by-step explanation:

Let father's age be 'x' and

Daughter's age be 'y'.

Now,

Two years ago father's age was 3.5times his daughter.

So,

x-2 = 3.5(y-2)

x-2 = 3.5y - 7

X = 3.5y - 5------1

Now,

after 6 years, father is twice his daughter's age plus 10 more.

Hence,the equation is :-

X+6 = 2(y+6)+10

substitute'X' value,we get :-

3.5y - 5 + 6 = 2y + 12 + 10

1.5y + 1 = 22

y = 21/1.5

y = 14----------2

Substitute'y' from 1

X = 3.5x14-5

X = 44

Hope it helps