Question

VW
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.
(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:

Age of father = 44 years

Age of daughter =14 years

Step-by-step explanation:

Let the present age of father be “x” years and the present age of daughter be “y” years.

A/q. 2 years ago,

x-2=3$\frac{1}{2}$*(y-2)

or  x-2=7/2(y-2)

or  2x-4=7(y-2)

or  2x-4 = 7y-14

⇒ 2x – 7y + 10 = 0 ……. (i)

A/q. 6 years hence,

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

or   x + 6 = 10 + 2y + 12

⇒ x – 2y – 16 = 0 ……. (ii)

Now, we multiply 2 to equation (ii) and then subtract eq.(i) from (ii):

2x – 4y – 32=0

-

2x – 7y + 10 =0

( -)      (+)       (-)

------------------------------

        3y-42=0

Now,3y=42

or y=14 years

Now, substituting the value of y = 14 in eq. (ii), we get

x – (2*14) – 16 = 0

⇒ x = 16 + 28

⇒ x = 44 years

Mark as the brainliest please  ^_^