Question

Two positive numbers x and y are such that x > y. If the difference of these numbers is 5 and their product is 24, find :
(i) sum of these numbers.
(ii) difference of their cubes.
(iii) sum of their cubes.​

Answer 1

Answer:

1)x=8

2)y=3

Step-by-step explanation:

1)×-y=5

2)xy=24

so,

×-y=5

x=5+y

apply x in 'xy=24'

xy=24

(y+5)×y=24. (x=5+y)

y×y+5y-24=0

y×y+(8-3)y-24=0

y×y+8y-3y-24=0

y(y+8)-3(y+8)=0

(y+8)(y-3)=0

y+8=0. y-3=0

y=-8 it's not possible. y=3

x-y=5

x-3=5

x=8

Answer 2

Answer:

Step-by-step explanation:

Two positive integers are x and y.

Difference of these numbers is 5.

x - y= 5

x=5+y-------♡

Product of these numbered is 24.

x×y=24

(5+y)×y=24

5y+y^2=24

y^2+5y-24=0

y^2-3y+8y-24=0(By splitting the middle term)

y(y-3)+8(y-3)=0

(y-3)(y+8)=0

y-3=0 or y+8=0

y=3 or y=-8

Since the given integers are positive,

y=3-------☆

From ♡ and ☆

x=5+3

x=8

Sum of these numbers=x+y

=8+3

=11.

Difference of their cubes =x^3-y^3

=8^3-3^3

=512-27

=485.

Sum of their cubes =x^3+y^3

=8^3+3^3

=512+27

=539.