Question

The product of the ages in years) of two sisters is 238. The difference in
their ages (in years) is 3. We would like to find their present ages.​

Answer 1

Answer:

17years and 14 Years.

Step-by-step explanation:

Let the Ages of the sisters be X and Y

Given XY = 238

⇒Y = 238 / X ...................(1)

Also , X - Y = 3 ................(2)

Substituting (1) in (2)

⇒X - $\frac{238}{X}$ = 3

⇒$\frac{X^{2} - 238 }{X} = 3$

⇒$X^{2} -238 = 3X$

⇒X² - 3X -238 = 0

⇒X² -17X + 14X - 238 = 0

⇒X(X -17) + 14 ( X - 17) = 0

⇒(X+14) (X-17) = 0

⇒iF,(X+14)= 0

     ⇒ X = -14             Not Posssible since age cannot be Negative.

⇒iF,(X-17)= 0

     ⇒ X = 17

Substituting X = 17 IN  (2)

X - Y = 3

⇒17 - Y = 3

⇒17 - 3= Y

⇒Y = 14

Hence the Present ages of the Sisters are 17years and 14 Years.

Thank U : )

Answer 2

Answer:

the present age of the sisters are 17 and 14 respectively

Step-by-step explanation:

let the age of 1 sister be x

let the age of the other one be y

according to question,

xy=238----------(1)

x+y=3------------(2)

x=y-3

put x=y-4 in equation 1

(y-3)y=238

y²-3y-238=0

y²-17y+14y-238

y(y-17)+14(y-17)

(y+14)(y-17)

y=-14 or y=17

as age cannot be negative so we'll discard -14

therefore, y=17

putting y=17 in equation 1

x*17=238

x=238/17

x=14