Question

willing to give medals and become if somebody could help me answer this question

Answer 1

Solve 6 over x minus 6 equals the quantity of x over x minus 6, minus six halfs for x and determine if the solutions is extraneous or not. for x and determine if the solution is extraneous or not.

Answer 2

here are the answer choices
x = -6, extraneous

x = -6, non-extraneous

x = 6, non-extraneous

x = 6, extraneous

Answer 3

Ur first step would be making all the fractions have a common denominator

To this multiply the first and second each with 2/2

Multiply the last one by (x-6)/(x-6)

Answer 4

ok well the first fraction would be 2x/2x-12 right @nikato

Answer 5

Yes, that's one of them

Answer 6

and would the second fraction be 6/x-6

Answer 7

right

Answer 8

No.

This is getting confusing..

6 2. x. 2. 6. (x-6)
---- * ----- = ------ *. ------ - ---- * -----
(x-6). 2. (x-6). 2. 2. (x-6)

Does this make sense? If so, can you simplify the numerator, because the denominator is all the same

Answer 9

yea kinda

Answer 10

So basically, I mutinied the first and second by 2/2 and the lat by (x-6)/(x-6)

Answer 11

*multiplied.

And now all 3 fractions share a common denominator, 2(x-6)

Answer 12

ok

Answer 13

Since we know our denominator is 2(x-6), what can't x be?

Remember the denominator can't =0

Answer 14

-6?

Answer 15

No, 6.
2(x-6)=0
---- --
2. 2

x-6=0
+6. +6
-----------
x=6

Answer 16

so its extraneous

Answer 17

Ok, we can just ignore the denominator now and work on the numerator.

6(2)= x(2) - 6(x-6)
Simplify and solve for x

Answer 18

ok i solved for it but was i right when i said it was extraneous @nikato

Answer 19

Oh ok. Yea. 6 is extraneous