anonymous
  • anonymous
What is the solution to the following rational equation?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[ \frac{ x }{ x ^{2} -9}-\frac{ 1 }{ x-3 }=\frac{ 1 }{ 4x-12 }\]
anonymous
  • anonymous
@ash2326 @hba @Hero @nincompoop @phi @satellite73
anonymous
  • anonymous
@Hero

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phi
  • phi
Can you factor each of the denominators ?
anonymous
  • anonymous
I know how to factor the first one... (x+3)(x-3)
anonymous
  • anonymous
the second one can't be and the last one I think is x-3
phi
  • phi
the last one you can factor 4 from each term. you get 4(x-3) (notice if you "distribute" the 4 you get 4x -12 so you know that is correct)
anonymous
  • anonymous
okay
phi
  • phi
now to solve, we could find a common denominator. Or, which is almost the same thing, is "clear the denominator" so far you have \[ \frac{x}{(x-3)(x+3)} - \frac{1}{(x-3)} = \frac{1}{4(x-3)} \] multiply both sides and all terms by (x-3) can you do that ?
phi
  • phi
you put (x-3) next to each term . It multiplies the top of each fraction notice you can cancel when the same thing is in the top and bottom
anonymous
  • anonymous
I don't understand...
phi
  • phi
multiply both sides of the equation by (x-3). That means multiply *every term* by (x-3) to multiply, write (x-3) next to each term, like this \[(x-3)\cdot \frac{x}{(x-3)(x+3)} - (x-3)\cdot\frac{1}{(x-3)} = (x-3)\cdot\frac{1}{4(x-3)} \]
phi
  • phi
you can think of (x-3) as \[ \frac{(x-3)}{1} \] and the (x-3) multiplies the top but notice you have \[ \frac{(x-3)}{(x-3)} \] in each term. This is the same as 1 (anything divided by itself is 1) so you can "cancel" x-3 from the top and bottom
phi
  • phi
what do you get ?
anonymous
  • anonymous
1?
phi
  • phi
\[ (x-3)\cdot \frac{x}{(x-3)(x+3)} - (x-3)\cdot\frac{1}{(x-3)} = (x-3)\cdot\frac{1}{4(x-3)} \] what is the first term simplify to?
anonymous
  • anonymous
x/(x+3)
phi
  • phi
yes. now the next term ?
anonymous
  • anonymous
do I distribute the negative?
phi
  • phi
first cancel, then distribute the -1
anonymous
  • anonymous
okay, so it would be -1
phi
  • phi
yes now that last term
anonymous
  • anonymous
1/4
phi
  • phi
so what do we now have ? \[ \frac{x}{x+3} - 1 = \frac{1}{4} \] we could do a few different things, but "combining like terms" is always a good step Can you add +1 to both sides and then simplify ?
anonymous
  • anonymous
I don't know how... -1 doesn't have a like term
phi
  • phi
the like terms are the "pure numbers" you know you can add two numbers together. so add +1 to both sides, like this \[ \frac{x}{x+3} - 1 +1 = \frac{1}{4} +1\] can you simplify ?
anonymous
  • anonymous
\[\frac{ x }{ x+3 }=1\frac{ 1 }{ 4 }\]
phi
  • phi
can you make 1 and 1/4 an "improper fraction" ?
anonymous
  • anonymous
5/4?
phi
  • phi
yes you now have \[ \frac{x}{x+3} = \frac{5}{4} \] any ideas what to try next ?
anonymous
  • anonymous
no
phi
  • phi
to solve these things you should have some ideas on what to do next. one thing you could say to yourself is " (x+3) in the denominator is ugly" If I multiply both sides of the equation by (x+3), it will cancel in the first term. try multiplying both terms by (x+3) what do you get ?
phi
  • phi
to multiply, write (x+3) next to each term
anonymous
  • anonymous
\[x=\frac{ 5 }{ 4 }\]?
anonymous
  • anonymous
@phi
phi
  • phi
Did you write (x+3) time each term in \[ \frac{x}{x+3} = \frac{5}{4} \] there are two terms (one on the left side of =, and the other on the right side. write (x+3) next to each term. can you do that ?
anonymous
  • anonymous
no...
phi
  • phi
just write (x+3) next to each term.
anonymous
  • anonymous
\[\frac{ x(x+3) }{( x+3)(x+3)}=\frac{ 5(x+3) }{ 4(x+3) }\]
phi
  • phi
that is correct (you multiplied each term by 1). but not what we want to do. just multiply each term by (x+3) (NOT (x+3)/(x+3) )
anonymous
  • anonymous
\[\frac{x(x+3) }{ x+3 }=\frac{ 5(x+3) }{ 4 }\]
phi
  • phi
yes. now notice on the left side of =, you have (x+3)/(x+3) you can "cancel" those. what do you get ?
phi
  • phi
do you agree that (x+3)/(x+3) is 1 ? so \[ \frac{x(x+3)}{(x+3) } = x \cdot 1 = x \]
phi
  • phi
you now have \[ x = \frac{ 5(x+3) }{ 4 } \] I would multiply both sides by 4. can you do that ?
phi
  • phi
you are almost done with this problem.

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