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solve fractions w/ variables: {x/(x^2-64)} + {8/(x-8)}...
- Thread starter anm2007
- Start date
D
Deleted member 4993
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Re: solve fractions with variables
anm2007 said:{ x / (x^2-64)} + {8 / (x-8)} = {2 / (x+8)}
how was my bracketing this time? good? thx.
Very good - Now show your work telling us exactly where you are stuck
You start off similar to your previous problem:
http://www.freemathhelp.com/forum/viewtopic.php?t=26552
Re: solve fractions with variables
ok so i factored out the denominator of the first set( x^2-64) and i got (x +8)(x-8)
then i to get both denominators the same i multiplied (x+8) to the top and bottom of the second set like this {8(x+8) / (x-8)(x+8)}...right?
Subhotosh Khan said:anm2007 said:{ x / (x^2-64)} + {8 / (x-8)} = {2 / (x+8)}
how was my bracketing this time? good? thx.
Very good - Now show your work telling us exactly where you are stuck
You start off similar to your previous problem:
http://www.freemathhelp.com/forum/viewtopic.php?t=26552
ok so i factored out the denominator of the first set( x^2-64) and i got (x +8)(x-8)
then i to get both denominators the same i multiplied (x+8) to the top and bottom of the second set like this {8(x+8) / (x-8)(x+8)}...right?
D
Deleted member 4993
Guest
Re: solve fractions with variables
anm2007 said:ok so i factored out the denominator of the first set( x^2-64) and i got (x +8)(x-8)
then i to get both denominators the same i multiplied (x+8) to the top and bottom of the second set like this {8(x+8) / (x-8)(x+8)}...right?
Correct and do similar thing on the right hand side so that everything is divided by (x^2-64). Then simplify ....
Re: solve fractions with variables
u mean on the other side of the equal sign? {1(x-8)/ (x+8)(x-8)}?Subhotosh Khan said:anm2007 said:ok so i factored out the denominator of the first set( x^2-64) and i got (x +8)(x-8)
then i to get both denominators the same i multiplied (x+8) to the top and bottom of the second set like this {8(x+8) / (x-8)(x+8)}...right?
Correct and do similar thing on the right hand side so that everything is divided by (x^2-64). Then simplify ....
D
Deleted member 4993
Guest
Re: solve fractions with variables
anm2007 said:u mean on the other side of the equal sign? {1(x-8)/ (x+8)(x-8)}?
Correct
Re: solve fractions w/ variables: {x/(x^2-64)} + {8/(x-8)}..
VERY NICE, anm.
But no need to go overboard; this is sufficient:
x / (x^2 - 64) + 8 / (x - 8) = 2 / (x + 8)
BUT there's nothing wrong with the way you did it;
appearance wise that is: easier to "read".
anm2007 said:{ x / (x^2-64)} + {8 / (x-8)} = {2 / (x+8)}
how was my bracketing this time? good? thx.
VERY NICE, anm.
But no need to go overboard; this is sufficient:
x / (x^2 - 64) + 8 / (x - 8) = 2 / (x + 8)
BUT there's nothing wrong with the way you did it;
appearance wise that is: easier to "read".
Re: solve fractions with variables
well then now i got {(9x+64) / (x-8)} / {(x+8)(x-8)} = {(1/ (x-8)}
i dont think this is right
Subhotosh Khan said:anm2007 said:u mean on the other side of the equal sign? {1(x-8)/ (x+8)(x-8)}?
Correct
well then now i got {(9x+64) / (x-8)} / {(x+8)(x-8)} = {(1/ (x-8)}
i dont think this is right
D
Deleted member 4993
Guest
\(\displaystyle \frac{x}{x^2-64} + \frac{8}{x-8} = \frac{2}{x+8}\)
\(\displaystyle \frac{x}{x^2-64} + \frac{8(x+8)}{x^2-64} = \frac{2(x-8)}{x^2-64}\)
\(\displaystyle \frac{9x + 64}{x^2-64} = \frac{2x-16}{x^2-64}\)
Since the denominetors are equal - numerators must be equal (for x not equal to ±8)
\(\displaystyle 9x + 64 = 2x - 16\)
Now continue....
\(\displaystyle \frac{x}{x^2-64} + \frac{8(x+8)}{x^2-64} = \frac{2(x-8)}{x^2-64}\)
\(\displaystyle \frac{9x + 64}{x^2-64} = \frac{2x-16}{x^2-64}\)
Since the denominetors are equal - numerators must be equal (for x not equal to ±8)
\(\displaystyle 9x + 64 = 2x - 16\)
Now continue....
Subhotosh Khan said:\(\displaystyle \frac{x}{x^2-64} + \frac{8}{x-8} = \frac{2}{x+8}\)
\(\displaystyle \frac{x}{x^2-64} + \frac{8(x+8)}{x^2-64} = \frac{2(x-8)}{x^2-64}\)
\(\displaystyle \frac{9x + 64}{x^2-64} = \frac{2x-16}{x^2-64}\)
Since the denominetors are equal - numerators must be equal (for x not equal to ±8)
\(\displaystyle 9x + 64 = 2x - 16\)
Now continue....
sorry i just realized that i wrote the question wrong ...on the other side of the equal sign its not {2/(x+8)} its {1/(x+8)}
D
Deleted member 4993
Guest
anm2007 said:Subhotosh Khan said:\(\displaystyle \frac{x}{x^2-64} + \frac{8}{x-8} = \frac{2}{x+8}\)
\(\displaystyle \frac{x}{x^2-64} + \frac{8(x+8)}{x^2-64} = \frac{2(x-8)}{x^2-64}\)
\(\displaystyle \frac{9x + 64}{x^2-64} = \frac{2x-16}{x^2-64}\)
Since the denominetors are equal - numerators must be equal (for x not equal to ±8)
\(\displaystyle 9x + 64 = 2x - 16\)
Now continue....
sorry i just realized that i wrote the question wrong ...on the other side of the equal sign its not {2/(x+8)} its {1/(x+8)}
Okay - so fix up the shown work accordingly and continue...
Subhotosh Khan said:anm2007 said:[quote="Subhotosh Khan":3u6oihbv]\(\displaystyle \frac{x}{x^2-64} + \frac{8}{x-8} = \frac{2}{x+8}\)
\(\displaystyle \frac{x}{x^2-64} + \frac{8(x+8)}{x^2-64} = \frac{2(x-8)}{x^2-64}\)
\(\displaystyle \frac{9x + 64}{x^2-64} = \frac{2x-16}{x^2-64}\)
Since the denominetors are equal - numerators must be equal (for x not equal to ±8)
\(\displaystyle 9x + 64 = 2x - 16\)
Now continue....
sorry i just realized that i wrote the question wrong ...on the other side of the equal sign its not {2/(x+8)} its {1/(x+8)}
Okay - so fix up the shown work accordingly and continue...[/quote:3u6oihbv]
ok i got -9 as my answer please tell me that is correct
D
Deleted member 4993
Guest
Put your answer back into left-hand-side of your given and simplify.
Put your answer back into right-hand-side of your given and simplify.
If those two numbers are equal - you are most probably correct.
Put your answer back into right-hand-side of your given and simplify.
If those two numbers are equal - you are most probably correct.
Subhotosh Khan said:Put your answer back into left-hand-side of your given and simplify.
Put your answer back into right-hand-side of your given and simplify.
If those two numbers are equal - you are most probably correct.
yes they checked out ...thank u so much for ur help. as u can tell i am not good at math at all...and it doesnt help that my teacher mumbles and talks to the board while "teaching" so i appreciate it.