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The sum of the ages of two brothers is 22 years.Six years ago the product of their ages was 21 years find the age of the elder brother

User Qiau
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1 Answer

23 votes
23 votes

Explanation:

x = age of brother 1 today

y = she of brother 2 today

x + y = 22

(x - 6)(y - 6) = 21

as 6 years ago, both were 6 years younger. than today.

from the first equation we get e.g.

x = 22 - y

using that in the second equation we get

(22 - y - 6)(y - 6) = 21

(16 - y)(y - 6) = 21

16y - 96 - y² + 6y = 21

-y² + 22y - 117 = 0

the general solution to a quadratic equation is

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

x = y

a = -1

b = 22

c = -117

y = (-22 ± sqrt(22² - 4×-1×-117))/(2×-1) =

= (-22 ± sqrt(16))/-2 = (-22 ± 4)/-2

y1 = (-22 + 4)/-2 = 9

y2 = (-22 - 4)/-2 = 13

as we can see, when we try it in our equations, x and y are interchangable : one has to be 9 and the other 13.

so, the older brother is 13 now.

User Aditya Pawade
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