Polynomial equations: sum is 10, product is 20; find sum of reci
- Thread starter eskimoxo
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D
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Re: Polynomial equations
Remember
\(\displaystyle \frac{1}{x} \, + \, \frac{1}{y} \, = \, \frac{x \, + \, y}{x\cdot y} \,\)
eskimoxo said:
Remember
\(\displaystyle \frac{1}{x} \, + \, \frac{1}{y} \, = \, \frac{x \, + \, y}{x\cdot y} \,\)
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How are "x" and "y" defined? Where do you use the fact that the sum of the two numbers is 10? Where do you use the equation, provided by the tutor, relating the denominators of two summed unitary fractions (fractions with "1" for their numerators) to the sums and products of those denominators?eskimoxo said:
Please reply with a clear listing of all of your work and reasoning. Thank you!
Eliz.
Re: Polynomial equations
i am not really sure how to write math equations in these boxes but i will try.
ok so the original question i am trying to solve is:
The sum of two numbers is 10. Their product is 20. Find the sum of their reciprocals.
my reasoning was that the equation should be xy + x + y = 10 + 20
then --> xy + x + y - 30 = 0
like i said before i tried to find to factor it but i could find anything.
so the the tutor gave me the equation using the reciprocals
1/x + 1/y = x+y/xy
when i simplify this i get 2x + 2y - xy = 0
and if you simplify x+y/xy= 10/20
= 1/2
so.... i don't know where to go from here.
i am not really sure how to write math equations in these boxes but i will try.
ok so the original question i am trying to solve is:
The sum of two numbers is 10. Their product is 20. Find the sum of their reciprocals.
my reasoning was that the equation should be xy + x + y = 10 + 20
then --> xy + x + y - 30 = 0
like i said before i tried to find to factor it but i could find anything.
so the the tutor gave me the equation using the reciprocals
1/x + 1/y = x+y/xy
when i simplify this i get 2x + 2y - xy = 0
and if you simplify x+y/xy= 10/20
= 1/2
so.... i don't know where to go from here.
D
Deleted member 4993
Guest
Re: Polynomial equations
If the numbers are 'x' and 'y' - then..;
\(\displaystyle \frac{1}{x} \, + \, \frac{1}{y} \, = \, \frac{x \, + \, y}{x\cdot y} \, = \, \frac{10}{20} \, = \, \frac{1}{2}\)
Subhotosh Khan said:
eskimoxo said:
If the numbers are 'x' and 'y' - then..;
\(\displaystyle \frac{1}{x} \, + \, \frac{1}{y} \, = \, \frac{x \, + \, y}{x\cdot y} \, = \, \frac{10}{20} \, = \, \frac{1}{2}\)
Re: Polynomial equations
1/x + 1/y = x+y/xy
1/x + 1/y = 10/20
1/x + 1/y = 1/2
then i multiplied everything by two to eliminate the half
2/x + 2/y = 1
then to get a common factor with the denominator
2y/xy + 2x/xy = 1xy/xy
2y+2x/xy = 1 <--- i cancelled the xy from this fraction
then multiply both sides by xy
2y + 2x = xy
2x -xy +2y = 0
1/x + 1/y = x+y/xy
1/x + 1/y = 10/20
1/x + 1/y = 1/2
then i multiplied everything by two to eliminate the half
2/x + 2/y = 1
then to get a common factor with the denominator
2y/xy + 2x/xy = 1xy/xy
2y+2x/xy = 1 <--- i cancelled the xy from this fraction
then multiply both sides by xy
2y + 2x = xy
2x -xy +2y = 0
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To refresh your memory on how to safely and clearly format math in a text-only environment, try reviewing the articles in the links you saw in the "Read Before Posting" thread that you'd read before originally posting. :idea:eskimoxo said:
Thank you!
Eliz.
mmm4444bot
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Factoring with respect to X and with respect to Y ...
Hello Eskimo:
I would answer your question by saying, "not in the way that you're probably thinking".
These forms don't apply to your exercise:
-[x(y - 2) - 2y] = 0
-[y(x - 2) - 2x] = 0
Cheers,
~ Mark
eskimoxo said:
Hello Eskimo:
I would answer your question by saying, "not in the way that you're probably thinking".
These forms don't apply to your exercise:
-[x(y - 2) - 2y] = 0
-[y(x - 2) - 2x] = 0
Cheers,
~ Mark