Question

2x+4/radical x+2

Which is equivalent to this?

Does this equal to 2 radical x+2?

If so, can someone demonstrate why?

Answer 1

\[\frac{2x+4}{\sqrt{x+2}}\]?

Answer 2

yes

Answer 3

\[\frac{2(x+2)}{\sqrt{x+2}}\]

Answer 4

is a start

Answer 5

and it is always the case that
\[\frac{z}{\sqrt{z}}=\sqrt z\]
you can check that easily by
1) squaring both sides and see that you get \(z\) or
2) write in exponential notation and subtract the exponents

Answer 6

so yes,
\[\frac{2(x+2)}{\sqrt{x+2}}=2\sqrt{x+2}\]

Answer 7

is that a specific formula?

Answer 8

no it is not a formula, it is just a fact

Answer 9

\[\frac{9}{\sqrt9}=\frac{9}{3}=3=\sqrt9\] for example

Answer 10

Alright I get that now. Mind if you help me with an inverse problem real quick? No one answered me correctly when I asked it.

Answer 11

\[\frac{x}{\sqrt{x}}=\frac{x}{x^{\frac{1}{2}}}=x^{1-\frac{1}{2}}=x^{\frac{1}{2}}=\sqrt{x}\]

Answer 12

y=3x+2/4x-7

Answer 13

i can try

Answer 14

LOL no wonder no one wanted to do it, is a bunch of pain in the butt algebra
solve
\[x=\frac{3y+2}{4y-7}\] for \(y\) takes a bunch of steps

Answer 15

we can walk through them if you like

Answer 16

i would love to

Answer 17

1) multiply both sides by \(4y-7\) to clear the denominator
let me know what you get

Answer 18

yeah thats what they told me before but i dont necessarily get why it clears the denominator i thought the conjugate does

Answer 19

you are not rationalizing a denominator
you are solving
if you had
\[\frac{3y+2}{4y-7}=5\] and had to solve for \(y\) what would you do?

Answer 20

make the fraction a whole number

Answer 21

what do you mean by that?
can you solve this \[\frac{3y+2}{4y-7}=5\]
?

Answer 22

how about this
\[\frac{2x}{3}=7\] what would you do here?

Answer 23

i dont know if anything i almost have no clue in is taking out denominators

Answer 24

ok how about this
\[\frac{x}{3}=5\]

Answer 25

multiply 5 by 3 and get
\[x=5\times 3=15\]

Answer 26

\[\frac{2x+1}{3}=5\\
2x+1=3\times 5=15\\
2x=14\\
x=7\]

Answer 27

oh....

Answer 28

first step is always to multiply by the denominator so you can solve the equation
\[\frac{x+1}{x-3}=7\\
x+1=7(x-3)\\
x+1=7x-21\\
etc\]

Answer 29

ok i get that now

Answer 30

want to try this one \[\frac{3y+2}{4y-7}=5\]

Answer 31

3y+2=20y-35

-33=17y

Answer 32

i mean 37=17y

Answer 33

ok now we get to the real question \[\frac{3y+2}{4y-7}=x\] which is harder because we have x and not 5, but it is the same steps exactly
multiply by \(4y-7\) to get it all on one line

Answer 34

what do you get?

Answer 35

3y+2=4xy-7x

Answer 36

shouldn't you switch the variables first?

Answer 37

ok good
now we are solving for \(y\) so we need all terms with \(y\) on one side of the equal sign, everything else on the other
yes, we switched already

Answer 38

\[3y+2=4xy-7x\] your choice as to where you want the y terms, right or left, makes no difference pick one

Answer 39

how do you minus the 4xy part again?

Answer 40

you want them on the left i guess
you just write it

Answer 41

\[3y+2=4xy-7x\\
3y+2-4xy=-7x\]

Answer 42

we got three steps to go
since you put the y terms on the left, put everything without a y in it on the right

Answer 43

well thats done right

Answer 44

uh no

Answer 45

you have a 2 on the left still

Answer 46

couldn't you just divide everything by -7 to get x now

Answer 47

we are solving for y
no you do not divide until the very last step

Answer 48

ok so 3y-4xy=-7x-2

Answer 49

yes

Answer 50

now we are solving for \(y\) and you have the two terms with \(y\) on one side of the equal sign
factor it out

Answer 51

you know what i mean by that?

Answer 52

yea but i dont know how to do it

Answer 53

\[3y-4xy=-7x-2\\
(3-4x)y=-7x-2\]

Answer 54

and now the LAST step is to divide to get \(y\)
\[y=\frac{-7x-2}{3-4x}\]

Answer 55

which i would probably write as
\[y=\frac{7x+2}{4x-3}\]

Answer 56

i think i just died a little

Answer 57

you want to do it again?

Answer 58

or we could do this one \[\frac{3y+2}{4y-7}=5\] again and see that the steps were identical

Answer 59

i think the main problem is how to cancel the denominator.

if you look at this other problem i asked http://openstudy.com/users/dtan5457#/updates/547d3062e4b0ae92adfff782

why did they mutliply by the conjugate, and the denominator goes away in that situation?

Answer 60

OK THANKS I ACTUALLY GOT IT NOW