Question

Solve 4 over x minus 4 equals the quantity of x over x minus 4, minus four thirds for x and determine if the solution is extraneous or not.
x = 4, extraneous
x = 4, non-extraneous
x = -4, extraneous
x = -4, non-extraneous

Answer 1

\[\frac{4}{x-4}=\frac{x}{x-4}-\frac{4}{3}\]?

Answer 2

yes

Answer 3

there is no solution, so not sure why you are supposed to come up with one

Answer 4

i guess since there is no solution you are supposed to say
x = 4, extraneous

Answer 5

if you subtract you get
\[\frac{4-x}{x-4}=-\frac{4}{3}\]

Answer 6

since \(\frac{4-x}{x-4}=-1\) you have
\[-1=-\frac{4}{3}\] which is silly

Answer 7

not sure why you are supposed to say "4, extraneous" when "no solution" should suffice, but what do i know? must be some math teacher way to solve that gives \(4\) as an answer, even though it isn't one

Answer 8

im not understanding

Answer 9

subtract \(\frac{x}{x-4}\) from both sides of the equation

Answer 10

how do people come up with these questions??!

Answer 11

you get
\[\frac{4-x}{x-4}=-\frac{3}{4}\]

Answer 12

the left hand side is a number, it is \(-1\)

Answer 13

the answer choices are under the question

Answer 14

@phi no clue

Answer 15

go with
x = 4, extraneous

Answer 16

extraneous is math teacherese for "no solution"

Answer 17

ok ill let you know if its right or wrong i have 51 questions so this is just the many ones XD

Answer 18

51 in 60 minutes lol

Answer 19

huh

Answer 20

if you clear the denominator (multiply by x-4 )
you can "solve" to get x=4

but really sats way of solving shows no solution with no extraneous value.

Answer 21

\(\large \dfrac{4}{x-4}=\dfrac{x}{x-4}-\dfrac{4}{3} \)

\(\large 3(x - 4)\dfrac{4}{x-4}=3(x - 4)\dfrac{x}{x-4}-3(x - 4)\dfrac{4}{3} \)

\(\large 12 = 3x -12x + 48\)

\(\large 9x = 36\)

\(\large x = 4\)

Since x = 4 cause a zero in the denominator, x = 4 is discarded and there is no solution.

Answer 22

ok thank you <3