## anonymous one year ago 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

1. anonymous

$\frac{4}{x-4}=\frac{x}{x-4}-\frac{4}{3}$?

2. anonymous

yes

3. anonymous

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

4. anonymous

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

5. anonymous

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

6. anonymous

since $$\frac{4-x}{x-4}=-1$$ you have $-1=-\frac{4}{3}$ which is silly

7. anonymous

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

8. anonymous

im not understanding

9. anonymous

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

10. phi

how do people come up with these questions??!

11. anonymous

you get $\frac{4-x}{x-4}=-\frac{3}{4}$

12. anonymous

the left hand side is a number, it is $$-1$$

13. anonymous

the answer choices are under the question

14. anonymous

@phi no clue

15. anonymous

go with x = 4, extraneous

16. anonymous

extraneous is math teacherese for "no solution"

17. anonymous

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

18. anonymous

51 in 60 minutes lol

19. anonymous

huh

20. phi

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.

21. mathstudent55

$$\large \dfrac{4}{x-4}=\dfrac{x}{x-4}-\dfrac{4}{3}$$ $$\large 3(x - 4)\dfrac{4}{x-4}=3(x - 4)\dfrac{x}{x-4}-3(x - 4)\dfrac{4}{3}$$ $$\large 12 = 3x -12x + 48$$ $$\large 9x = 36$$ $$\large x = 4$$ Since x = 4 cause a zero in the denominator, x = 4 is discarded and there is no solution.

22. anonymous

ok thank you <3