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 twicethe age of his daughter then. Find their present ages.(1) Let the present age of the father be x years and that of his daughter be y years.(2) Form two equations from the given conditions.(3) Solve the equations and find the answer.​

Answer 1

Answer:

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

$x - 2 = 3 \frac{1}{2} (y - 2) \\ x - 2 = \frac{7}{2} (y - 2) \\ x - 2 = \frac{7y}{2} - 7 \\ x - \frac{7y}{2} + 5 = 0 \\ \frac{2x - 7y + 10}{2} = 0 \\ 2x - 7y + 10 = 0 \\ 2x - 7y = - 10 \: \: \: \: \: \: equ \: (1)$

6 years later

father age = x + 6

daughters age = y + 6

father will be 10 years more than twice of daughters age

$x + 6 = 2(y + 6) + 10 \\ x + 6 = 2y + 12 + 10 \\ x + 6 = 2y + 22 \\ x - 2y - 16 = 0 \\ x - 2y = 16\: \: \: \: equ \: (2)$

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