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Yes!

What's your question?

O_o question? XD 2+2=?

Not allowed to give direct answers. Check the drawing:
|dw:1433043776037:dw|

4 LOL

3 +_+

gimme a minute!

Calculate \(\large \frac{1}{2}+\frac{2}{3}\). Show work.

7/6

Calculate 12+23. \(\huge \color{red}{Show work.}\)

Why are we doing this? .-.

Because it is the same way you do this as with rational fractions.
This is a pretest!

\[12+23=35\]

idk how to show work for that .-.

2+1=3
3+2=5

ok, that's how it works:
Calculate \(\large \frac{1}{2}+\frac{2}{3}\), show work.

\(\large \frac{1}{2}+\frac{2}{3}=\frac{1*3}{2*3}+\frac{2*2}{3*2}\)

Do you know why we multiply the first fraction by three, and the second by two?

to get a common denominator

Exactly!

Just a review of terms:
1/2 = 2/4 because they are __________________

equivalent

exactly "equivalent fractions"! That's a good start! You know your stuff.

:)

The idea of adding and subtracting rational fraction is exactly the same!

Since each term remains an equivalent fraction, the answer will not change, BUT...

we now only have to add the numerators of the equivalent fractions, just like the numerical example.

Sorry it took a while!

this looks easy :)

The next one will be equally easy!
gimme a minute.

Calculate \(\large \frac{1}{x}+\frac{2}{x^2}\) show work!

\[\huge~\frac{ 1*x }{ x*x }+\frac{ 2 }{ x^2}=\frac{ 1x+2 }{ x^2 }\]

Excellent! Just remember that 1x is usually written as x, except when showing work.

Now:
Calculate \(\large \frac{2}{(x-1)}-\frac{1}{(x-1)^2}\) show work!

\[\frac{ 2*(x-1) }{ (x-1)*(x-1) }-\frac{ 1 }{ (x-1)^2 }=\frac{ 1(x-1) }{ (x-1) }=1\]

T_T not allowed!

;-;

i dont get it

Good! I'll explain!

\(\large \frac{ 2*(x-1) }{ (x-1)*(x-1) }-\frac{ 1 }{ (x-1)^2 }\)
is perfect!

following so far?

yes

Now can you do the math (on the numerator) and finish?

\[\frac{ x(x-1) }{ (x-1)^2 }\]

You need to distribute the first term, and add like terms,
2*(x-1)-1 = 2x-2 -1 =2x-3

following?

Yes

So the final answer is....

\[\huge\frac{ x(x-1) }{ 2x-3 }\]

>_<

wrong spot...

\[\frac{ 2x-3 }{ (x-1)^2 }\]

Whew!
Yes, excellent!

:/ so its easier to multiply or is it to break down the numbers? :/

yes

Are there any points to clarify, or are we good?

We're good.

ok, now, try this:
Calculate \(\large \frac{3}{x}-\frac{2}{x-1}\) show work!

ummm O_O

Look at the denominators.

* 2*5 in denominator

O_o

i thought u can only do that if there is an equal sighn :/

It's not a real cross multiplication, but it works in a similar way.

The pattern helps when there are no common factors.
You can use that in the new problem.

:?

hmm ok makes a bit more sense

Good!
you can finish it?

\[\frac{ x-1 }{ x(x-1) }\]

brb

With x-3 as numerator, what would the answer be then?

>_< idk :(

The first part is very good!

O_O

x_x

seems easier... :)

Good!
Would you like to do one while I'm offline, probably in less than 5 minutes?

Sure i gotta go too xD

\(\large \frac{3}{x-1}-\frac{2x}{x^2-1}\)

huh? O_o

http://prntscr.com/7bnxzi

Simplify \(\large \frac{3}{x-1}-\frac{2x}{x^2-1}\). Show work!

im not sure how to do this .-.

\[\frac{3*x+1 }{ x-1*x+1 }-\frac{ 2x }{ (x+1)(x-1) }=\frac{ 6x(x+1) }{ (x^2-1 }\]

Ew factoring

no wait

i was gonna say that XD

of course! I knew that! xD

lol u are learning all mai tricks >.>

Well, this is either called learning, or contamination, depending on the point of view! xD

contamination >.> because now i cant use it :P

xD

Just another contamination!

LOL

Ok, would you feel sad if I say...

Are we done for the day? :o

Not at all

nope, but we're done with 11.2 to 11.6 ?

We are done? :o YAY NO MORE MATH

Not too fast, read the whole line, please! lol

So now we're onto Rational equations! Interesting stuff!

T_T

Solve for x: \(\large x+1=\frac{72}{x}\)

i dont think i have learned this .-.

If not, now you will!

O_O

wait im starting to remeber things >.>

Any idea where to start?

\[\huge~\frac{ x+1 }{ 1}=\frac{ 72 }{ x }\]

Excellent! Memories are coming back!

next?

\[ \huge~72*1=x(x+1)\]
\[\huge~72=x^2+1x\]

Excellent again!
next?

\[\huge~x^2+1x-72=0\]

Beautiful! Pray continue!

|dw:1433116854133:dw|
\[\huge~(x+9)(x-8)\]

You're almost done. Equate each factor to zero to find the answers.

x=-9
x=8

100%
Excellent job!

I think i have this section lets not do anymore :D ?

Just two more!
Solve for x: \(\large \frac{10}{x+4}=\frac{15}{4(x+1)}\)
This is easier than FOIL.

NO T_T

There is no factoring involved, you'd do it in 10 seconds, + 30 minutes of typing.

\[\huge~40(x+1)=15(x+4)\]

Too fast.... but everything is right!

40x+40=15x+60
40x=15x+20
25x=20
x=20/25

Excellent, and the answer is.....

x=20/25

... or 4/5

OR 20/25.

smh

xD

Now the last one, you'd be sad when it's done!

WAIT

After this no more math??? :O

T_T <---tears of joy

Yeah, after this last problem, no more math (until your quiz) unless you have questions.

i have 2 questions on my hw thing i dont get at all : /

\[\huge~\frac{ 3x }{ 1 }=\frac{ 5x+6 }{ 2x+3 }\]
\[\huge3x(2x+3)=1(5x+6)\]

Excellent, so far so good. You're living up to your reputations!

\[6x^2+9x=5x+6\]
\[6x^2+4x-6=0\]
|dw:1433118120725:dw|

So far so good! Pray continue!

|dw:1433118297396:dw|

None.

Excellent. What's the next step?

idk

Quadratic formula!

OMG NO T_T

So much work

lets just skip this one ;-; lets do one that doesnt have this

What would the equation be?

:/

a=3
b=2
c=-3
\[\huge~x=\frac{ -2 \pm \sqrt{(-2)^2-4(3)(-3)} }{ 2(3) } \]

Correct! Now pull out your calculator, or do it mentally, to find the final solution.

\[\huge~x=\frac{ -2\pm~\sqrt{40} }{ 6 }\]

\[\huge~x=\frac{ -2 \pm \sqrt{4\times10} }{ 6 }\]

\[\huge~x=\frac{ -2 \pm 2\sqrt{10} }{ 6 }\]

T_T

:)

O_O where did u get that?

Does it look too difficult?
BTW, the answer is -3, so you don't have to cheat! lol

noo it looked fimiliar

omg no way T_T

hmmmmk

You said you had two hw questions!

nvm :)

Can we practice more on everything once yr back? :)

sure!

Ready. :)

Start with 11.2?

or you have specific questions?

No lets just go with 11.2 etc

|dw:1433125010350:dw|
name the variation.

inverse variation

Good!
x 1 2 3 4 5
y 2 4 6 8 10
what is x when y=14

7? :/

Yep, good!

x 1 5 10
y 20 4 2
What is y when x=40

80

Hmmm....

10?

We see that y is decreasing as x increases. See if you can find the rule!

0

2

Another way to write the rule is xy=k
For example 2, k=20,
so 1*20=20 4*5=20...

x 2 7 12
y 6 21 36
Name the variation, and find x when y=120

i dont like there >.>

these

You've done those before?

I know we did not go over it, but never too late! :(

|dw:1433125957505:dw|

Notice that inverse variation is not a straight line. It is a curve.

ok so far?

Yea

Ok, now, to find the rule, this is what you do.

First find if the function is increasing, or decreasing.
ok

do you know how to tell if a function is increasing?

If the line is going up?

|dw:1433126223857:dw|

Yes. An excellent example.
If it is a straight line, we know it is not inverse.

Right :)

Is it automatically direct variation?

yes? :/

Lets skip this.

|dw:1433126603203:dw|

hmmmm makes a bit more sense :3

T_T
120=k/x

Can you start with the variation? (is it increasing? If it is, it is not inverse)

increasing .-.

so it cannot be inverse. The rule you gave is for inverse.

so y=mx+b?

When you give a rule, you have to check that it works for at least two points.

mm ;-; <---

40

mmm k

Now, try
x 2 4 6
y 6 3 2
State
1. the variation
2. the rule
3. value of x when y=24

1) PArtial
2) y=mx+b
3) Lets skip it ;-;

Well, it's decreasing

so it is not direct.
Is it a straight line, let's check the slope.

So it's not partial.
Try inverse:
x 2 4 6
y 6 3 2
2*6=12
4*3=12
6*2=12
so k=12 for inverse!

Can you finish answering?

y=12x :/

Variation: inverse
Rule: either xy=12, or y=12/x
Now when y=24, what would x be?

2

Hin:
Rule: either xy=12, or y=12/x

weve spent about an hour on this .-.
y=2? ??

xy=12, y=24
so x=12/y=12/24=1/2 lol
Yes, almost an hour.

and it was a good thing, because now you won't miss the easy questions.

k, where do we go from here?
(it's laggy)

Want to start a new post?

ok