Question

I am in desperated need for some algebra 2 help!! I will fan and medal anyone who helps me I take a test tomorrow and I have no idea what i'm doing!!

Answer 1

What is the solution to the rational equation x over 2x minus 1 plus 1 over 4 equals 2 over 2x minus 1 ?

x = 9 over 2

x = 7 over 2

x = 3 over 2

x = 7 over 6

Answer 2

\[\large \frac{x}{2x - 1} + \frac{1}{4} = \frac{2}{2x - 1}\] like that?

Answer 3

Yes, that's what it looks like.

Answer 4

Are you able to help me? @johnweldon1993

Answer 5

I'd certainly like to think so... Multiply everything by \(\large 2x - 1\)

\[\large 2x - 1 \times \frac{x}{2x - 1} + 2x-1 \times \frac{1}{4} = 2x - 1\frac{2}{2x - 1}\]

Notice

\[\large \cancel{2x - 1} \times \frac{x}{\cancel{2x - 1}} + 2x-1 \times \frac{1}{4} = \cancel{2x - 1}\frac{2}{\cancel{2x - 1}}\]
so all we have left is
\[\large x + \frac{2x-1}{4} = 2\]

look better? what would you do next?

Answer 6

yes and umm i'm not really sure this stuff confuses me.. @johnweldon1993

Answer 7

Well, did what I do confuse you?

Answer 8

No, I got that part I just don't know what to do next.

Answer 9

Okay good, well that was the worst part...

now

\[\large x + \frac{2x - 1}{4} = 2\]

If we multiply everything by 4...we would have

\[\large 4x + \cancel{4}\times \frac{2x -1}{\cancel{4}} = 8\]

\[\large 4x + 2x - 1 = 8\]

what next?

Answer 10

Then you'd combine like terms

Answer 11

Correct....which would leave you with...?

Answer 12

6x-1=8

Answer 13

Now just solve for 'x'

Answer 14

x=9/6

Answer 15

which then reduces to 3/2

Answer 16

And that is your answer! :)

Answer 17

Can you help me with come more? I'd really appreciate it!

Answer 18

Of course :)

Answer 19

Want me to open a new question or just do it on this one?

Answer 20

Hmm just do it here, no need to open a new one :)

Answer 21

What is the simplified form of the quantity 4 x squared minus 25 over the quantity 2 x plus 5 ?

2x - 5, with the restriction x ≠ -five over 2

2x + 5, with the restriction x ≠ five over 2

2x - 5, with the restriction x ≠ five over 2

2x + 5, with the restriction x ≠ -five over 2

Answer 22

\[\large \frac{4x^2 - 25}{2x + 5}\]

So right here....a restriction would be when the denominator = 0...because when the denominator = 0 we have an undefined function...so

when does

\[\large 2x + 5 = 0\]?

Answer 23

you'd get x=-5/2

Answer 24

Correct! That is your restriction

Answer 25

Ohhhh!

Answer 26

Now, we just need to simplify the expression

Answer 27

\[\large \frac{4x^2 - 25}{2x + 5}\]

So...how can we simplify that top part?

Answer 28

(4x-5)(x+5)?

Answer 29

Very close! however when you expand that back out you get

4x^2 + 20x - 5x - 25

4x^2 + 15x - 25

so not quite....any other thoughts?

Answer 30

umm no not really I thought that was the right way to simplify that

Answer 31

well remember,4 has 2 different ways to factor...we can have 4 and 1.....OR you can do 2 and 2 right?

Answer 32

Yes

Answer 33

Soooo, lets try out

(2x + 5)(2x - 5)

expand that out..what do we get?

Answer 34

4x^2-10x+10x-25

Answer 35

mmhmm, and combine like terms... we get back to the original 4x^2 - 25 right?

Answer 36

yes

Answer 37

It's always good to check if you can get back to the start...

so now we have

\[\large \frac{(2x - 5)(2x + 5)}{2x + 5}\]

Well notice...
\[\large \frac{(2x - 5)\cancel{(2x + 5)}}{\cancel{2x + 5}}\]

so all we're left with is

\[\large 2x - 5\]

Answer 38

ahhh its all making sense now.

Answer 39

Good :) remember, ask anything that doesn't make sense. Might as well ask while I'm here :P

Answer 40

What is the solution to the rational equation x over x squared minus 9 plus 1 over x minus 3 equals 1 over 4x minus 12 ?

Answer 41

You have been a ton of help I really appreciate it!

Answer 42

\[\large \frac{x}{x^2 - 9} + \frac{1}{x - 3} = \frac{1}{4x - 12}\]

any idea where to start with this one?

Answer 43

Well I would start by I guess trying to like cancel out the bottom terms

Answer 44

Right....go ahead and let me know when you get stuck :)

Answer 45

\[\frac{ x }{ x+3 }+\frac{ 1 }{ x-3 }=\frac{ 1 }{4}\]
thats what I got so far

Answer 46

wait...where did that come from?