Question

see if i am correct

Answer 1

\[\frac{ 2x }{ x +4 } -\frac{ x }{ x-4}\]

Answer 2

you flip the second fraction which make is look like this \[\frac{ x-4 }{ x}\]

Answer 3

\[\frac{ 2x }{ x+4 } -\frac{ x-4 }{ x}\]

Answer 4

@phi

Answer 5

not correct

Answer 6

how am i not correct?

Answer 7

well so far

Answer 8

you don't flip when adding or subtracting. (when you divide, you "flip")

if you had 2/4 of a pizza and eat ¼
2/4 - ¼ you do (2-1)/4 = ¼ left. No flipping!

Answer 9

ohh ok

Answer 10

when you see fractions that are added (or subtracted) think: COMMON DENOMINATOR

in this case (x+4)(x-4) is the common denominator.
Next, think: how do I make the first fraction have a bottom of (x+4)(x-4) ?

Answer 11

idk

Answer 12

@phi

Answer 13

the first fraction already has (x+4) in its bottom
how do you make it have (x+4)(x-4) ? multiply it by (x-4)

but you are allowed to do that only if you do the same thing to the top
so multiply the first fraction by (x-4)/(x-4)
can you do that ?

Answer 14

ok so i guess you multiple this fraction right \[\frac{ 2x }{ x+4 } \times \frac{ x-4 }{ x-4 }\]

Answer 15

well it should look like that

Answer 16

yes, but leave the bottom as (x+4)(x-4)
the top becomes 2x^2 - 8x

Answer 17

\[ \frac{ 2x^2-8x }{ (x +4)(x-4) } -\frac{ x }{ x-4} \]
now for the other fraction, what do we multiply the bottom by so that we get
(x+4)(x-4) ?
and then do the same to the top.

Answer 18

x(x+4) right?

Answer 19

you multiply the top and bottom by (x+4)

Answer 20

or is it x^2+4x

Answer 21

and ik you do the mutliple the bottim but right now im focused on doing the top right now

Answer 22

*multiple

Answer 23

you now have
\[ \frac{ 2x^2-8x }{ (x +4)(x-4) } -\frac{ x^2+4x }{ (x +4)(x-4) } \]

Answer 24

now the fractions have the same bottom, so we can combine them
\[ \frac{2x^2 -8x - (x^2+4x)}{(x+4)(x-4) } \]
notice I put in parens because we subtract the whole thing (x^2 +4x)

Answer 25

oh ok i get it thanks

Answer 26

do you think you could help me with three more

Answer 27

first, did you finish this one. the top simplifies.

Answer 28

oh wait for the prob we just did what are the restriction that exist with (or in) the resulting rational expression.

Answer 29

yes, there are restrictions.

Answer 30

oh for the top part of the fraction divide everything by 2 right

Answer 31

ok wouldnt the final answer be \[x^{2}-4x-x ^{2}+2x\]

Answer 32

over (x+4)(x-4)

Answer 33

looking just at the top
\[ 2x^2 -8x - (x^2+4x) \]
or
\[ 2x^2 -8x + -1\cdot (x^2+4x) \]
the minus one can be "distributed"
\[ 2x^2 -8x + -1x^2 -1\cdot 4x \]

the -1 * 4x is -4x, so this is
\[ 2x^2 -8x + -1x^2 -4x \]
we have a 2x^2 - 1x^2 (2 x squareds take away one x squared )
that is 1x^2 or just x^2
\[ x^2 -8x -4x\]
can you combine the -8 x - 4 x ?

Answer 34

yes

Answer 35

it would be -12x

Answer 36

@phi

Answer 37

yes, so the top is x^2 -12x
and the answer is
\[ \frac{x^2-12x}{(x-4)(x+4)}
\]
or (if you like... it does not matter which way we write it)
\[ \frac{x(x-12)}{(x-4)(x+4)}
\]

what x values will cause a *divide by zero* ?

Answer 38

zero?

Answer 39

you will divide by 0 if (x-4) is 0
does that make sense ?

Answer 40

ohh no it doesnt

Answer 41

if (x-4) is 0 we will end up dividing by 0

Answer 42

now does this have to deal with the restriction in the problem

Answer 43

x-4 = 0 when x = 4
because 4-4 is 0

if x is 4, we will be doing this problem:
\[ \frac{4(4-12)}{(4-4)(4+4)} \]
or
\[ \frac{4(4-12)}{(0 \cdot 8)} \]
or
\[ \frac{4(4-12)}{0} \]

we are not allowed to divide by 0, so we must not allow x to be 4

Answer 44

ohhhhh ok now i got you

Answer 45

there is another "bad" x value. what is it ?

Answer 46

the bottom is (x-4)(x+4)
we know x=4 makes (x-4) be zero, and we don't want that.
What about (x+4) ? could some x value make that 0 ?