Question

........

Answer 1

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

Answer 2

@Napolions

Answer 3

a=2 and b=3

Answer 4

what up

Answer 5

can i get a picture

Answer 6

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

Answer 7

alright so which part u need elp with 1 or 2

Answer 8

help

Answer 9

both...

Answer 10

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

Answer 11

alright so do you want to find the variable x

Answer 12

is it 4?

Answer 13

yes nice job

Answer 14

it was 3/4

Answer 15

wat was 3/4?

Answer 16

the answer it was 3 over 4

Answer 17

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

Answer 18

2+x =3 and 2x =4

Answer 19

call my name when you need help

Answer 20

wait how did u get that???

Answer 21

@Napolions

Answer 22

what up

Answer 23

adding 1 up top and multiplying 2 at the bottem

Answer 24

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

Answer 25

which is \[x+2=6\]

Answer 26

now your confusing me

Answer 27

@Tazmaniadevil

Answer 28

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

Answer 29

yes

Answer 30

so thats the first part right?

Answer 31

yes

Answer 32

ok wat about the second part???

Answer 33

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.

Answer 34

that is how you identify it

Answer 35

@plzzhelpme

Answer 36

i didnt get that @Napolions

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

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

Answer 41

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

Answer 42

cross multiplication??

Answer 43

think so

Answer 44

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

Answer 45

@Napolions

Answer 46

what up

Answer 47

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

Answer 48

x_x

Answer 49

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)

Answer 50

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

Answer 51

ok so make the x 1

Answer 52

it was
x+22x=3x

3(2x)=x(x+2)

6x=x^2+2x


\[x^2+2x-\left(6x\right)=6x-\left(6x\right)\]

Answer 53

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

Answer 54

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

Answer 55

\[x=\frac{-\left(-4\right)+\sqrt{\left(-4\right)^2-4\cdot \:0\cdot \:1}}{2\cdot \:1}=4\]and
\[x=\frac{-\left(-4\right)-\sqrt{\left(-4\right)^2-4\cdot \:0\cdot \:1}}{2\cdot \:1}=0\]

Answer 56

can u check it once??

Answer 57

ok

Answer 58

\[x=4,\:x=0\]
right...?

and x=0 is the extraneous

Answer 59

correct

Answer 60

oh ok thxxxx.......

Answer 61

Medal plzzz

Answer 62

u know i did most of the question ^.^

Answer 63

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

Answer 64

yeah but i helped though but yeah you did most

Answer 65

lol