Math, asked by joonswurld, 16 days ago

four years ago a father was six times as old as his son. ten years hence age of father will be 5/2 times the age of son determine the present age of father and son. solve this in elimination method

Answers

Answered by mimopro19
0

Answer:

40 and 10

Step-by-step explanation:

let the present age of the father be x and that of the son be y,

taking the case of four years ago,

x-4 = 6(y-4)

x-4 = 6y-24

x = 6y-20 ...(1)

taking the case of ten years in future,

x+10 = \frac{5}{2}(y+10)

2(x+10) = 5(y+10)

2x + 20 = 5y + 50

2x = 5y + 30...(2)

now using the elimination method on eq(1) and eq(2) by multiplying eq(1) by 2 and subtracting eq(2) from eq(1),

2x - 2x = 12y - 5y -40 - 30

0 = 7y - 70

y = \frac{70}{7}

y = 10 (substitute this value in eq(1))

x = 6(10) - 20\\x = 60 - 20\\x = 40

so the father's current age is 40 and the son's current age is 10

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