Question

five year ago a man was seven times as old as his son . five year hence ,the father will be three times as old as his son . find their present ages​

Answer 1

Answer:

Let the present ages of son and father be x and y respectively

five years ago,

man's age =y-5

son's age=x-5

A/Q, y-5=7(x-5)

y-5=7x-35

7x-35=y-5

7x-y-30=0....(1)

5years hence,

man's age=y+5

son's age =x+5

y+5=3(x+5)

y+5=3x+15

3x+15=y+5

3x-y+10=0...…(2)

solve both equations by elimination method,

7x-y-30=0

3x-y+10=0

4x-40=0

4x=40

x=10

put this value in eq^n (1)

7*10-y-30=0

40-y=0

y=40

Hence, man's age= 40years

son's age=10 years

Hope this helps

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Answer 2

$\large\underline\pink{Given:-}$

• five year ago a man was seven times as old as his son .
• five year hence ,the father will be three times as old as his son .

$\large\underline\pink{To find:-}$

• find their present ages. ....?

$\large\underline\pink{Solutions:-}$

Five years age a ma. was seven times as old as his son.

• let the age of son be x years age of gather = 7x

so,

• present age of son = x + 5
• present age of father = 7x + 5

five year, the father will be three to times as old as his son.

• present age of son = x + 5 + 5 = x + 10
• present age of father = 7x + 10

so, the father will be three times as old as his son.

⟹ 7x + 10 = 3(x + 10)

⟹ 7x + 10 = 3x + 30

⟹ 7x - 3x = 30 - 10

⟹ 4x = 20

⟹ x = 5

Now, present age of father

⟹ 7x + 5

⟹ 7 × 5 + 5

⟹ 35 + 5

⟹ 40 year's.

present age of son

⟹ x + 5

⟹ 5 + 5

⟹ 10 year's.

Hence,

• present age of father = 40 year's
• present age of son = 10 year's