please please please help
2007-12-02 02:38:52 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
19x + 47y = 1000
19x = 1000 - 47y
y = 14, x = 18 give an integral pair
1000 = 47*14 + 19*18
**EXTRA**
19x = 1000 - 47y
= (19*52 + 12) - 57y + 10y
x = 52 - 3y + (12 + 10y)/19
So we require 12 + 10y = 19*k
or 10y = 19k - 12
We require 19k to be a multiple of 19 ending in 2. So k must end in 8. That narrows it down considerably.
So k could be 8, 18, 28, 38, 48.
The respective values of y are:
14, 33, 52, 71, 90
The respective values of x are:
18, -29, -76, -123, -170
So if you're interested in positive integers, then
x=18, y=14 is the solution pair.
2007-12-02 02:43:08 · answer #1 · answered by Dr D 7 · 4⤊ 1⤋
Make a system of equations.
Answer: 342 and 658
(342/19 = 18 and 658/47 = 14)
(342+658=1000)
2007-12-02 02:52:49 · answer #2 · answered by snakker2k 6 · 0⤊ 0⤋
you can't. The prime factors of 1000 are 2,2,2,5,5,5 When multiplied together the answer is 1000.
Also 19 and 47 are both prime numbers. The only way a number can be divided into sub parts like your question asks is if 19 AND 47 are factors of the original number. For example the number 2679 has 3 factors 19, 47, 3. There is no answer to your question.
2007-12-02 02:47:03 · answer #3 · answered by Rich 7 · 0⤊ 2⤋
19x + 47y = 1000
19(18) + 47(14) = 1000
342 + 658 = 1000
2007-12-02 02:55:43 · answer #4 · answered by fed up woman 6 · 0⤊ 1⤋
658 and 342
2007-12-02 02:45:55 · answer #5 · answered by Anonymous · 0⤊ 1⤋
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2016-10-25 07:38:58 · answer #6 · answered by desmangles 4 · 0⤊ 0⤋
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2007-12-02 02:40:44 · answer #7 · answered by Anonymous · 0⤊ 3⤋