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@Nnesha its not spam ;)
@Nnesha its not spam ;)

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Other answers:

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
That means: 1. i dont like this and 2. i dont understand this OR 3. i dont want to do this Which one/ones?
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!
We multiply top and bottom of each term by a factor so that the denominator becomes the common denominator.
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.
\(\large \frac{1}{2}+\frac{2}{3}=\frac{1*3}{2*3}+\frac{2*2}{3*2}=\frac{3}{6}+\frac{4}{6}=\frac{7}{6}\)
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!
\[\frac{ 2*(x-1) }{ (x-1)*(x-1) }-\frac{ 1 }{ (x-1)^2 }=\frac{ 1(x-1) }{ (x-1) }=1\]
The denominator is still meant to be (x-1)^2 ? You need to do the math on the numerators. Give it another shot!
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!
At this point, both denominators are identical, so we can do the math just on the numerators \(\large \frac{ 2*(x-1) -1 }{ (x-1)^2 }\)
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!
One thing we did without saying it is...nothing. We did "nothing" to factorize the numerator, because it was "obvious" there are no factors.
In general, after the arithmetic on the numerator, we need to factor the numerator, can you tell me why?
:/ so its easier to multiply or is it to break down the numbers? :/
So that we can cancel (with condition) IF there are common factors between the numerator and denominator, just like what we did before in the simplification. Does that make sense?
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.
there are no common factors! So you just "cross multiply", which is a short cut when there are no common factors. like \(\large \frac{1}{2}+\frac{3}{5}= \frac{1*5}{2*5}+\frac{2*3}{x*5}=\frac{1*5+2*3}{2*5}=\frac{11}{10}\)
* 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.
Calculate \(\large \frac{3}{x}-\frac{2}{x-1}\), show work! Remember that when there is no common factors in the denominators, the common denominator is just the product, namely x(x-1)
:?
\(\large \frac{3}{x}-\frac{2}{x-1}=\frac{3(x-1)}{x(x-1)}-\frac{2x}{(x-1)x}=\frac{3(x-1)-2x}{x(x-1)}\)
hmm ok makes a bit more sense
Good! you can finish it?
\[\frac{ x-1 }{ x(x-1) }\]
brb
IF it was x(x-1)/(x-1), you would have simplified it to x, x\(\ne\)1. But 3(x−1)−2x=3x-3 -2x (distribute 3(x-1) to 3x-3 first, then add -2x) =x-3
With x-3 as numerator, what would the answer be then?
>_< idk :(
the common denominator does not change, so still x(x-1) Answer would then be: \(\huge \frac{ x-3 }{ x(x-1) }\) nothing to further simplify!
Here's one for you to think about when I'm off for a few minutes. Calculate \(\large \frac{2}{(x+1)}+\frac{1}{(x-1)}\) , show work! Do remember to work out the math in the numerator, and the common denominator (product of x+1 and x-1) does not change.
\[\frac{ (x-1)*2 }{ (x-1)*(x+1) }+\frac{ (x+1)*1 }{ (x+1)*x-1) }=\frac{ 3(x+1)(x-1) }{ (x-1)(x+1) } \]
The first part is very good!
Recall that since the denominator is the same, and does not change, we need to ADD the numerators (and not multiply).
So it would read: \(\LARGE \frac{ (x-1)*2 + (x+1)*1}{ (x-1)(x+1) }\) \(=\LARGE \frac{ 2X-2 + X+1}{ (x-1)(x+1) }\) \(=\LARGE \frac{ 3X-1 }{ (x-1)(x+1) }\), OR \(=\LARGE \frac{ 3X-1 }{ x^2-1 }\)
O_O
Your steps are correct. It's just the last step of expanding and adding that there was a correction to make. Are you ok with the correction, or confused?
x_x
Well, if it's the part of expansion perhaps you could use some help. (x-1)*2 + (x+1)*1 is the same as 2(x-1) + (x+1) .... a coefficient of 1 is understood. Then we distribute, basically a half FOIL 2*x + 2*(-1) + x+1 ..... parentheses after a plus sign can be removed without change that makes 2x -2 +x +1 Now add like terms 2x+x -2+1 =3x-1 this is the numerator, since the denominator does not change, we have as the answer: =\(\Large \frac{ 3X-1 }{ x^2-1 }\)
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 .-.
I would factor the second denominator, x^2-1 into (x+1)(x-1) using difference of 2 squares (note 1=1^2). This gives \(\large \frac{3}{x-1}-\frac{2x}{(x+1)(x-1)}\) the proceed as before.
\[\frac{3*x+1 }{ x-1*x+1 }-\frac{ 2x }{ (x+1)(x-1) }=\frac{ 6x(x+1) }{ (x^2-1 }\]
A minor correction by adding parentheses: (using PEMBAS) \(\Large\frac{3*(x+1) }{ (x-1)*(x+1) }-\frac{ 2x }{ (x+1)(x-1) }=\frac{ 3x+3-2x }{ (x^2-1) }=?\)
The first step was perfect. Then you would distribute(expand) and add/subtract as needed. The final step I leave to you, if possible.
Ew factoring
no wait
in this case, there is nothing to factor! 3x-3-2x = 3x-2x-3=x-3 so answer is \(\Large \frac{x-3}{x^2-1}\)
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
Spam!Spam!Spam!Spam!Spam!Spam!Spam!Spam!Spam!Spam!Spam! Spam!Spam!Spam!Spam!Spam!Spam!Spam!Spam!Spam!Spam!Spam!
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.
I think most teachers would tolerate this (not simplifying), although I don't. I would give 95% instead of 100%. So it depends on your teacher!
smh
You're in trouble when I learn more of these! Well, you need to give extra credit to those who do a perfect job, that's my reasoning!
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 : /
We'll look at that after this question. Perhaps by that time you will know how to do. Solve \(\large 3x=\frac{5x+6}{2x+3}\). Show work.
\[\huge~\frac{ 3x }{ 1 }=\frac{ 5x+6 }{ 2x+3 }\] \[\huge3x(2x+3)=1(5x+6)\]
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!
Actually, to make you life simpler, divide the equation by 2 to get 3x^2+2x-3=0 This will give smaller numbers, so easier to work with.
|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
The reason I choose this one is because in the real world (including exams), answers don't always come out in integers.
Just to make sure you're prepared for the quadratic formula. You could type the discriminant faster than lightning, why not the quad. formula?
What would the equation be?
:/
\(6x^2+4x-6=0\), or half of that: 3x^2+2x-3 Either one will do, doesn't make a differce with the quad. formula. Makes a big diff. with factoring.
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.
\[\huge~x=\frac{ -2\pm~\sqrt{40} }{ 6 }\]
\[\huge~x=\frac{ -2 \pm \sqrt{4\times10} }{ 6 }\]
\[\huge~x=\frac{ -2 \pm 2\sqrt{10} }{ 6 }\]
Yes, yours is correct (again 95%), unless ... \(\Large~x=\frac{ -2 \pm \sqrt{40} }{ 2(3) }\) \(\Large~x=\frac{ -1 \pm \sqrt{10} }{ 3 }\)
T_T
lol, don't worry about the 5%, but I give you credit for coming that close, especially the part going from sqrt(40) to sqrt(10), excellent job!
:)
Well, in case you have nightmares with + and - signs poking at you, you can wake up and try this. It's not as hard as it seems. \(\Large \frac{3c}{c^2-4}+\frac{1}{c-2}=\frac{2}{c+2}\) You will need to simplify the left-hand side first, then do the equation solving.
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
Did you get it from your teacher? Well, I'll show you a gold mine. If you can't sleep, go to http://www.regentsprep.org/regents/math/algtrig/ate11/RationalEqPract.htm and practise all those problems!
omg no way T_T
With this one, do the simplification on the left-hand side and then solve. Do not follow the solution provided.
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
no, that's too small. The rule for direct variation is y=kx, where k is a constant. In the first example, x 1 2 3 4 5 y 2 4 6 8 10 we see that the rule is y=2x, so it is a direct variation.
The rule for inverse variation is y=k/x. For the second example, we have x 1 5 10 y 20 4 2 so the rule is y=20/x (20=20/1, 4=20/5, 2=20/10, ...) so what is y when x=40?
2
Another way to write the rule is xy=k For example 2, k=20, so 1*20=20 4*5=20...
According to the rule, y=20/40 = 1/2, so when x=40, y=1/2 or use the alternate rule, xy=20 40y=20, or y=1/2
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! :(
I'll show a few examples first. Direct variation: rule y=kx (when x is greater, y is greater) Inverse variation: rule xy=k (when x increases, y decreases, that's why inverse) so far so good?
|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? :/
Unfortunately, there are other variations. The one you drew is called a partial variation. It is a straight line that does not pass through the origin. A straight line that passes through the origin AND increasing, then it's direct!
Lets skip this.
|dw:1433126603203:dw|
It has to do with the rule. A partial is y=mx+b where b\(\ne\)0 A direct is when y=kx (i.e. b=0) For example, x 2 5 8 y 6 12 18 is partial because y=2x+2 while x 2 5 8 y 4 10 16 is direct because y=2x
hmmmm makes a bit more sense :3
Good! We'll get back to the last problem. x 2 7 12 y 6 21 36 Name the variation, state the rule, and find x when y=120
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 ;-; <---
and you need to find k. Let's say we decide that it is direct variation, then y=kx substitute (2,6) we have 6=2k, or k=3 Let's try it for another point: (12,36), so that is 36=12k, k=3 again. So we have found the rule y=3x. If y=120, then 120=3x, what is x?
40
Exactly. So the rule is: 1. examine and assume a distribution. 2. substitute one of the points in the rule (of the assumed distribution) 3. check with another point, to see if the rule is still valid. 4. Give the answer.
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.
x 2 4 6 y 6 3 2 slope = (y2-y1)/(x2-x1) between 2 and 4, slope = (3-6)/(4-2) = -3/2 = -1.5 between 4 and 6, slope = (2-3)/(6-4) = 1/2 Since the slope is not constant, it is not a straight line.
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

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