anonymous
  • anonymous
........
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[\frac{ x+a }{ ax }=\frac{ b }{ x }\]
anonymous
  • anonymous
@Napolions
anonymous
  • anonymous
a=2 and b=3

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anonymous
  • anonymous
what up
anonymous
  • anonymous
can i get a picture
anonymous
  • anonymous
ther isnt a picture its just the equation.......
anonymous
  • anonymous
alright so which part u need elp with 1 or 2
anonymous
  • anonymous
help
anonymous
  • anonymous
both...
1 Attachment
anonymous
  • anonymous
\[\frac{ x+2 }{ 2x }=\frac{ 3 }{ x }\]
anonymous
  • anonymous
alright so do you want to find the variable x
anonymous
  • anonymous
is it 4?
anonymous
  • anonymous
yes nice job
anonymous
  • anonymous
it was 3/4
anonymous
  • anonymous
wat was 3/4?
anonymous
  • anonymous
the answer it was 3 over 4
anonymous
  • anonymous
how?....... i got 4.....
anonymous
  • anonymous
2+x =3 and 2x =4
anonymous
  • anonymous
call my name when you need help
anonymous
  • anonymous
wait how did u get that???
anonymous
  • anonymous
@Napolions
anonymous
  • anonymous
what up
anonymous
  • anonymous
adding 1 up top and multiplying 2 at the bottem
anonymous
  • anonymous
im confused......... so first u multiply both sides by 2x.......then it gets canceled...... leaving \[x+2=\frac{ 3 }{ x }2x\]
anonymous
  • anonymous
which is \[x+2=6\]
anonymous
  • anonymous
now your confusing me
anonymous
  • anonymous
@Tazmaniadevil
anonymous
  • anonymous
and then u subtract 2 on both sides and get 4 \[x=4\]
anonymous
  • anonymous
yes
anonymous
  • anonymous
so thats the first part right?
anonymous
  • anonymous
yes
anonymous
  • anonymous
ok wat about the second part???
anonymous
  • anonymous
An extraneous solution is a root of a transformed equation that is not a root of the original equation because it was excluded from the domain of the original equation.
anonymous
  • anonymous
that is how you identify it
anonymous
  • anonymous
@plzzhelpme
anonymous
  • anonymous
i didnt get that @Napolions
anonymous
  • anonymous
this website should help u Extraneous Solutions - Hotmath hotmath.com/hotmath_help/topics/extraneous-solutions.htm
anonymous
  • anonymous
i mean this one Extraneous Solutions - Hotmath hotmath.com/hotmath_help/topics/extraneous-solutions.htm
anonymous
  • anonymous
ya that page isnt opening...... @Napolions
anonymous
  • anonymous
go to google and type in extraneous solutions and it is under the one that say hot math
anonymous
  • anonymous
wait i think i did it wrong........ \[\frac{ x+2 }{ 2x } =\frac{ 3 }{ x }\] \[ 3(2x)= x(x+2)\] \[6x=2x+2x\] \[6x=4x\] @Napolions
anonymous
  • anonymous
cross multiplication??
anonymous
  • anonymous
think so
anonymous
  • anonymous
im totally confused rn.........
anonymous
  • anonymous
@Napolions
anonymous
  • anonymous
what up
anonymous
  • anonymous
wat do i doo???>.........
anonymous
  • anonymous
x_x
anonymous
  • anonymous
how does this sound??? x(x+1) = 2x x^2 + x = 2x x^2 -x = 0 x(x-1)=0 x=0 or x=1 the x=0 is extraneous, because we are not allowed to divide by 0 (which is what happens in the original equation if x is 0)
anonymous
  • anonymous
i made a mistake in my previous problem....i did x*x is 2x butits x^2
anonymous
  • anonymous
ok so make the x 1
anonymous
  • anonymous
it was x+22x=3x 3(2x)=x(x+2) 6x=x^2+2x \[x^2+2x-\left(6x\right)=6x-\left(6x\right)\]
anonymous
  • anonymous
\[x^2-4x=0\]
anonymous
  • anonymous
are u checking to see if you are right do you need my help
anonymous
  • anonymous
\[x=\frac{-\left(-4\right)+\sqrt{\left(-4\right)^2-4\cdot \:0\cdot \:1}}{2\cdot \:1}=4\]and \[x=\frac{-\left(-4\right)-\sqrt{\left(-4\right)^2-4\cdot \:0\cdot \:1}}{2\cdot \:1}=0\]
anonymous
  • anonymous
can u check it once??
anonymous
  • anonymous
ok
anonymous
  • anonymous
\[x=4,\:x=0\] right...? and x=0 is the extraneous
anonymous
  • anonymous
correct
anonymous
  • anonymous
oh ok thxxxx.......
anonymous
  • anonymous
Medal plzzz
anonymous
  • anonymous
u know i did most of the question ^.^
anonymous
  • anonymous
lol but watev thx..........
anonymous
  • anonymous
yeah but i helped though but yeah you did most
anonymous
  • anonymous
lol

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