math hw, have anyone help me ar

2008-10-07 10:16 pm

i have two math questions meet difficulty ,i hope have anyone help me slove those two math question , thank you so much
question1:The product of two integers is 108 and the sum of their reciprocals is 7/36. what are the integers?
question2: A two-digit number is equal to twice the product of the digits. It is 27 less than the number obtained by reversing its digits. Find the number.

above those two qustions need to write step ar, thanks a lot

回答 (1)

Uncle Michael

2008-10-07 10:45 pm

✔ 最佳答案

Question1:

Let y be one of the integers.
Then another integer is 108/y.

The sum of their reciprocals:
(1/y) + (y/108) = 7/36
Multiply both sides by 108y:
108(1/y) + 108(y/108) = 108(7/36)
108 + y2 = 21y
y2 - 21y + 108 = 0
(y - 12)(y - 9) = 0
y = 12 ooro y = 9
108/y = 9 ooro 108/y = 12

Ans: The integers are 9 and 12.

=====
Question2:

Let p and q be the tenths digit and unit digit respectively.

10p + q = 2pq ...... (1)
(10q + p) - (10p + q) = 27 ...... (2)

(2):
10q + p - 10p - q = 27
9q - 9p = 27
q - p = 3
q = p + 3 ...... (3)

Put (3) into (1):
10p + (p + 3) = 2p(p + 3)
10p + p + 3 = 2p2 + 6p
2p2 - 5p - 3 = 0
(p - 3)(2p + 1) = 0
p - 3 = 0 ooro 2p + 1 = 0
p = 3 ooro p = -1/2 (rejected)

Put p = 3 into (3):
q = (3) + 3
q = 6

Ans: The number is 36.
=

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