Expansions.
Two positive numbers x and y are such that x>y. If the difference of these nunbers is 5 and their product is 24,find :
(i) Sum of these numbers.
(ii) difference of their cubes.
(iii) sum of their cubes.
sol...
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x-y = 5
Therefore x = 5 + y ----------(transposing it on that side)
xy =24
Substitute x as (y+5)
(y+5)y = y² +5y =24
y² + 5y - 24 = 0
y² +8y -3y -24 = 0
y(y + 8) -3( y +8 ) = 0
(y-3)(y+8) = 0
y = 3 , -8
Since they are positive numbers y =3
x = y + 5 = 3 + 5 = 8
Therefore the two numbers are 3 and 8
therefore:
1) sum of numbers = 8+3 =11
2) diiference of their cubes = 8³ -3³ = 512 - 27 = 485
3) sum of their cubes = 8³ +3³ = 512+27 = 539
Hope this helped u!!
Cheers!!
Therefore x = 5 + y ----------(transposing it on that side)
xy =24
Substitute x as (y+5)
(y+5)y = y² +5y =24
y² + 5y - 24 = 0
y² +8y -3y -24 = 0
y(y + 8) -3( y +8 ) = 0
(y-3)(y+8) = 0
y = 3 , -8
Since they are positive numbers y =3
x = y + 5 = 3 + 5 = 8
Therefore the two numbers are 3 and 8
therefore:
1) sum of numbers = 8+3 =11
2) diiference of their cubes = 8³ -3³ = 512 - 27 = 485
3) sum of their cubes = 8³ +3³ = 512+27 = 539
Hope this helped u!!
Cheers!!
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