anonymous
  • anonymous
Please help me solve this and check for extraneous solutions. 10 over x+4=15 over 4x+4
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
@oaktree
OakTree
  • OakTree
Right here.
OakTree
  • OakTree
Is it this:\[\frac{ 10 }{ x+4 } = \frac{ 15 }{ 4x+4 }\]

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anonymous
  • anonymous
Yay, yes
anonymous
  • anonymous
@oaktree ?
anonymous
  • anonymous
"cross multiply" and get \[\frac{ 10 }{ x+4 } = \frac{ 15 }{ 4x+4 }\iff 10(4x+4)=15(x+4)\]
anonymous
  • anonymous
you good from there?
anonymous
  • anonymous
Okay 40x+40=15x+60
anonymous
  • anonymous
yeah three more steps
anonymous
  • anonymous
Yes I believe so. Thank you but what would I do with a problem like....x over x-2 plus 1over x-4=2 over x squared-6x+8
anonymous
  • anonymous
It has 3 instead of two @satellite73
anonymous
  • anonymous
i would try to write it so i can understand it is it \[\frac{x}{x-2}+\frac{1}{x-4}=\frac{2}{x^2-6x+8}\]
anonymous
  • anonymous
\[\frac{x}{x-2}+\frac{1}{x-4}=\frac{3}{x^2-6x+8}\]
anonymous
  • anonymous
that right?
anonymous
  • anonymous
It is the first one
anonymous
  • anonymous
add on the left and get \[\frac{x}{x-2}+\frac{1}{x-4}=\frac{2}{x^2-6x+8}\] \[\frac{x(x-4)+(x-2)}{(x-2)(x-4)}=\frac{2}{(x-2)(x-4)}\]
anonymous
  • anonymous
cooked up so that the denominator on the left is the same as the denominator on the right after you factor
anonymous
  • anonymous
since the denominators are the same, you can equate \[x(x-4)+x-2=2\]and solve the quadratic
anonymous
  • anonymous
ooooh you have to be real real careful here
anonymous
  • anonymous
when you solve the quadratic, one solution will be \(x=4\) but that is not a solution to the original equation, because the original expressions are undefined if \(x=4\) so omit that solution
anonymous
  • anonymous
What does that mean?
anonymous
  • anonymous
(Omit)
anonymous
  • anonymous
what does what mean?
anonymous
  • anonymous
Yes
anonymous
  • anonymous
oh, it means don't put that as a solution, because it is not one
anonymous
  • anonymous
The answer that I got is (x+1)(x+2)
anonymous
  • anonymous
Oh, thankyou
anonymous
  • anonymous
Is that wrong?
anonymous
  • anonymous
\[x(x-4)+x-2=2\] \[x^2-4x+x-2=2\] \[x^2-3x-4=0\] \[(x-4)(x+1)=0\]
anonymous
  • anonymous
solutions to this are \(x=4\) or \(x=-1\) but as i said, \(4\) is not a solution to the original equation as it would make the denominator zero
anonymous
  • anonymous
So then would I put both down answers?
anonymous
  • anonymous
no you put down \(x=-1\) only
anonymous
  • anonymous
Can you help me with one more that is different from that one?
anonymous
  • anonymous
@satellite73

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