katherinesmith
  • katherinesmith
solve for x : equation inside
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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katherinesmith
  • katherinesmith
\[\frac{ 3 }{ x + 2 } - \frac{ 1 }{ x } = \frac{ 1 }{ 5x }\]
katherinesmith
  • katherinesmith
phi: this is my last question of the night. hahaha :)
phi
  • phi
bring 1/5x to the left side find a common denominator

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katherinesmith
  • katherinesmith
i guess x is the common denominator?
katherinesmith
  • katherinesmith
\[\frac{ 3 }{ x + 2 } - \frac{ 1 }{ x } - \frac{ 1 }{ 5x } = 0\]
phi
  • phi
this is an equality, so not as tricky. another way to do this is multiply both sides (and all terms) by (x+2) and then by 5x (this "clears" the denominators) can you do that ?
katherinesmith
  • katherinesmith
how would that clear the middle denominator?
phi
  • phi
one step at at time. what do you get if you multiply all terms and both sides of the equation by (x+2) ?
katherinesmith
  • katherinesmith
i have no idea.
katherinesmith
  • katherinesmith
if you show me what it looks like when you multiply all the terms by (x + 2) i will do it and then try it myself with 5x
phi
  • phi
just write (x+2) next to each term
phi
  • phi
and remember you can simplify (x+2)/(x+2) to 1
katherinesmith
  • katherinesmith
\[3 - \frac{ 1 }{ x ( x + 2) } - \frac{ 1 }{ 5x ( x + 2) } = (x + 2)\]
katherinesmith
  • katherinesmith
actually its all equal to 0. dumb error in my fault
phi
  • phi
yes, and you should by multiplying by (x+2). that means (x+2) goes in the "top" you divided by (x+2) (except for the first term, which you did correctly)
katherinesmith
  • katherinesmith
\[3 - \frac{ x + 2 }{ x } - \frac{ x + 2 }{ 5x } = 0\]
phi
  • phi
now multiply all terms, both sides by 5x. write 5x (in the top) next to each term
katherinesmith
  • katherinesmith
\[3(5x) - \frac{ 5x ( x + 2) }{ x } - 1 = 0\]
phi
  • phi
the last term is not -1. just put 5x next to the last term.
katherinesmith
  • katherinesmith
my mistake. \[3(5x) - \frac{ 5x ( x + 2 }{ x } + 5x = 0\]
katherinesmith
  • katherinesmith
is that right?
phi
  • phi
5x times the last term. not just 5x
katherinesmith
  • katherinesmith
\[3 ( 5x) - \frac{ 5x ( x + 2) }{ x } - \frac{ 5x }{ 5x} = 0 \]
katherinesmith
  • katherinesmith
???
phi
  • phi
\[ 3(5x) - \frac{ 5x(x + 2) }{ x } - \frac{ 5x(x + 2) }{ 5x } = 0(5x) \] now simplify. If you have the samething in the top and bottom, you can cancel them
katherinesmith
  • katherinesmith
what do i do with \[\frac{ 5x ( x + 2) }{ x }\]
phi
  • phi
the x in the top and bottom cancel, because x/x = 1
katherinesmith
  • katherinesmith
so then i get 10x?
katherinesmith
  • katherinesmith
\[15x - 10x - x + 2 = 0\]
phi
  • phi
15x is ok \[ 3(5x) - \frac{ 5\cancel{x}(x + 2) }{ \cancel{x} } - \frac{ \cancel{5x}(x + 2) }{ \cancel{5x} } = 0(5x) \\ 15x - 5(x+2) -(x+2) = 0 \] can you finish ?
katherinesmith
  • katherinesmith
yes
katherinesmith
  • katherinesmith
9x = -12
katherinesmith
  • katherinesmith
x = -12/9
katherinesmith
  • katherinesmith
correct?
katherinesmith
  • katherinesmith
actually x = 12/9
phi
  • phi
what do you get when you distribute the -5 in -5(x+2) ? you should get -5 times each term
katherinesmith
  • katherinesmith
yeah, -5x + -10
katherinesmith
  • katherinesmith
15x - 5x - 10 - x - 2 = 0 9x - 12 = 0 9x = 12 x = 12/9
phi
  • phi
you can simplify. top and bottom can be divided by 3
katherinesmith
  • katherinesmith
x = 4/3

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