........

- anonymous

........

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- anonymous

\[\frac{ x+a }{ ax }=\frac{ b }{ x }\]

- anonymous

@Napolions

- anonymous

a=2 and b=3

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## More answers

- anonymous

what up

- anonymous

can i get a picture

- anonymous

ther isnt a picture its just the equation.......

- anonymous

alright so which part u need elp with 1 or 2

- anonymous

help

- anonymous

both...

##### 1 Attachment

- anonymous

\[\frac{ x+2 }{ 2x }=\frac{ 3 }{ x }\]

- anonymous

alright so do you want to find the variable x

- anonymous

is it 4?

- anonymous

yes nice job

- anonymous

it was 3/4

- anonymous

wat was 3/4?

- anonymous

the answer it was 3 over 4

- anonymous

how?....... i got 4.....

- anonymous

2+x =3 and 2x =4

- anonymous

call my name when you need help

- anonymous

wait how did u get that???

- anonymous

@Napolions

- anonymous

what up

- anonymous

adding 1 up top and multiplying 2 at the bottem

- anonymous

im confused.........
so first u multiply both sides by 2x.......then it gets canceled......
leaving \[x+2=\frac{ 3 }{ x }2x\]

- anonymous

which is \[x+2=6\]

- anonymous

now your confusing me

- anonymous

@Tazmaniadevil

- anonymous

and then u subtract 2 on both sides and get 4
\[x=4\]

- anonymous

yes

- anonymous

so thats the first part right?

- anonymous

yes

- anonymous

ok wat about the second part???

- 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

that is how you identify it

- anonymous

@plzzhelpme

- anonymous

i didnt get that @Napolions

- anonymous

this website should help u
Extraneous Solutions - Hotmath
hotmath.com/hotmath_help/topics/extraneous-solutions.htm

- anonymous

i mean this one
Extraneous Solutions - Hotmath
hotmath.com/hotmath_help/topics/extraneous-solutions.htm

- anonymous

ya that page isnt opening...... @Napolions

- anonymous

go to google and type in extraneous solutions and it is under the one that say hot math

- 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

cross multiplication??

- anonymous

think so

- anonymous

im totally confused rn.........

- anonymous

@Napolions

- anonymous

what up

- anonymous

wat do i doo???>.........

- anonymous

x_x

- 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

i made a mistake in my previous problem....i did x*x is 2x butits x^2

- anonymous

ok so make the x 1

- 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

\[x^2-4x=0\]

- anonymous

are u checking to see if you are right do you need my help

- 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

can u check it once??

- anonymous

ok

- anonymous

\[x=4,\:x=0\]
right...?
and x=0 is the extraneous

- anonymous

correct

- anonymous

oh ok thxxxx.......

- anonymous

Medal plzzz

- anonymous

u know i did most of the question ^.^

- anonymous

lol but watev thx..........

- anonymous

yeah but i helped though but yeah you did most

- anonymous

lol

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