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\(\dfrac{2x}{4}-\dfrac{3}{4}\quad\Longleftrightarrow\dfrac{2x-3}{4}\)
Hay there

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so then I went
\[\frac{ 2x-3 }{ 4 }=\frac{ 10-15x }{ 6 }\]
Yes, then you cross-multiply
\[12x-18=40-60x\]
\[72x-18=40\]
yeah
\[72x=58\]
Is that enough to tell what kind of equation it is?
you can tell that x is going to equal something, so it wont be indentity.
two other options are conditional equation and inconsistent equation
Inconsistant i believe?
conditional equation true for certain value of x, where if you simplify equation, you end up with x=4 or something inconsistent equation false equation, where you cannot solve for x.
Oh so conditional is just like a usual equation that works out ?
If you have something like x+1 = x+2, it's inconsistent equation because you end up with 1=2
yeah
lel wow the facepalms are coming out today boys
1c. 0=8 so this is false or inconsistent?
pretty much both
I am having many issues on 2a
inconsistent equation is kind of fancy way to say false equation lol
you would need to combine fraction on right side. To do so, you need to get denominator to be same
you have \(y-2\) and \(y+1\) So you just multiply numerator and denominator by \(y+1\) for first fraction and \(y-2\) for second
\[\dfrac{1}{y-2}-\dfrac{2}{y+1}\quad\longrightarrow\dfrac{y+1}{y+1}\cdot\dfrac{1}{y-2}-\dfrac{2}{y+1}\cdot\dfrac{y-2}{y-2}\]
Does that make sense? Basically you just need to get denominator to be same, so you just multiply numerator and denominator by something that would get you to desired denominator
kinda hard for me to explain lol
so 1/(y+1)(y-2) - 4/(y+1)(y-2) ?
\[\dfrac{y+1}{y+1}\cdot\dfrac{1}{y-2}-\dfrac{2}{y+1}\cdot\dfrac{y-2}{y-2}\\~\\= \dfrac{y+1}{(y+1)(y-2)} - \dfrac{2(y-2)}{(y+1)(y-2)}\\~\\=\dfrac{y+1-2(y-2)}{(y+1)(y-2)}\]
ok let's start with something simply. \(\dfrac{1}{6}+\dfrac{1}{2}\) How can we combine fraction here?
multiply by 3/3
so 1/6+1/2*3/3
so you would get 1/6+3/6
right, so for problem 2a. it's same idea, just with variable.
get what I am saying here?
Yes I do its just weird for me to visualize and see it right away t-t idk why
because it's something new for you, I guess?
Well like the bigness of the equation kinda like put me in my place kinda deal
ok, so now we have \[\dfrac{y+1-2(y-2)}{(y+1)(y-2)}\] simplify numerator
\[y+1-2y+4\]
CLEP test seem little harsh lol...
don't forgot combine like term
yeh was just showing steps
-y+5
It does seem harsh :/ I'm gonna like work myself to the point of no return so I do well on it . I want this
Right. Now cross multiply \[\dfrac{7}{y^2-y-2} = \dfrac{-y+5}{(y-2)(y+1)}\longrightarrow~?\]
so we have to do the same with the other side then get (y-2)(y+1) as denominator?
oh i guess we dont
Not really. denominator doesn't matter if they are on different side of equation.
\[y^2(-y+5) ,-y(-y+5),-2(-y+5)\] would this work?
so I got -y3 + 5y^2 y^2 + 5y 2y+10
before combining
-y^3 + 5y^2 y^2 - 5y 2y-10
damn i messed up multiplying
you just messed up with +/- signs
\[-y^2+6y^2-3y-10\]
yeah
now the other side
let me delete that i need to slow down
7(y+1) 7(y-2) 14y+7, 14y-14 14y+14y, 7-14
none. left side should start with \(7(y+1)(y-2)\)
so just keep it as is o.o?
So firstly, you expand \((y+1)(y-2)\)
why not work out?
You only distribute something with plus sign. \(a(b+c) = ab+ac\) But \(a(bc) = abc\)
Alrights
make sense?
Yes it does
y^2+-2y+y-2?
Yeah
so we have \[y^2-y-2=-y^3+6y^2-3y-10\]
you forgot 7
\[7(y^2-y-2)=-y^3+6y^2-3y-10\]
damn
so do we expand it out with 7 now?
yup
\[7y^2-7y-14=-y^3+6y^2-3y-10\]
I got that
now take all terms to right side; \(0 = -y^3-y^2+4y-4\)
can confirm got that after working out
lel sorry I doing this in a note book :p
know what to do from here?
Notice something you can factor out? Hint: \[0 = \boxed{-y^3-y^2}+\boxed{4y-4}\]
we can simplify ?
oh you asked lol
Oops should be \(0 = -y^3-y^2+4y+4\)
would we simplify the y's?
well, actually we try to factor, so we can easily figure out solutions.
-1(y^2+4)(y+1)
Would that be how we factor it?
close. should be \((y+1)(4-y^2)\)
Then you can factor \(4-y^2\)
or you can "factor out" -1 from \(4-y^2\) So you would have something like \((-1)(y^2-4)\)
-y^2+4. Doesnt this just put us back where we were?
Yeah, just to make it slightly easier to factor out, but never mind.
factoring -1 out is just optional.
You got \(0 = (y+1)(2-y)(y+2)\)?
Yes that would be it fully factored
yeah, now can you find solutions?
I was thinking, maybe we need to do something more efficient with time. we spend a lot of time on few problems lol.
indeed. Im just confusedwhere to go from here like.... you are dealing with 3 exponents of y and you need to get it to 1 y to solve for the question
you are confused on how to get solution?
Yes... like right now I feel like you could take -y^2 add to both sides and square both sides so you could get y=sqrt-y^3+4y-4
But that wouldn't work so I just am straight confused where to go from here
while solving for something, we want equation to be fully factored, so we can apply something (I forgot name) that states that If ab = 0, then either a = 0 or b = 0
So we can "split" equation into \(0 = y+1\) or \(0=2-y\) or \(0=y+2\)
So from here, just solve for y from each equation Does that make sense?
Who gave you medal? lol
Lol I have no clue, and Yeah that makes sense in order to get it fully factored when you have many parts
y=-1 or y=2 or y=-2
hmm Now plug y = -1 or y = 2 back into original equation
what happens?
i got 7/0=1/0-2/3
yeah. zero in denominator. So we are left with y = -2 That it lol
wait how we left with -2?
s: we still had -2/3 left ove
y = -1 and y = 2 get us zero in denominator, so they are invalid solutions.
oh that makes sense nvm so we would try y=-2
and baddabing there it comes out
You can. and you will get true equation.
Yeh Imma try that second one alone and check it laters just to make sure I understand this
alright
|dw:1434844521381:dw|
ehh not right reaction, but whatever
loool wowsers
Alright so with the first word problem., Ralph = 6-2n? Ralph + Tommy= 30?
R = 2T - 6 R+T = 30
Could we set the equation to 2t-6=30? Cause we know the total it could be
|dw:1434844843534:dw|
yeah you could do that
|dw:1434844913540:dw|
you have beautiful mouse control lol
thanks lol
3t=36 36/3 t=12
so we know Tommy scored 12 12*2 R=24-6=18
|dw:1434845125612:dw|
lol Loki pls
So this one is worded weirdly for me it sounds like we are using multiple variables?
OH wait no
twice the 3rd is 86
86/2
no
let's go ahead and set up equation
Alright
we can let first age be \(a\) So second age would be \(a+2\) third age is \(a+4\) Because we are given that three ages arethree even CONSECUTIVE integers
"If the sum of the 1st, four times the 2nd, and twice the 3rd is 86, what are the three ages?" So we have \(a + 4(a+2)+2(a+4) = 86\) Make sense?
Yep. As I fail reading it. It sounded really weird to me egh.
yeah weird worded, I had to read twice lol.
a+4a+8+2a+8=86 7a+16=86 7a=70 a=10 First Kid=10 Second Kid= 12 Third Kid = 14
Yep!
The next one is 20%*n=960?
or .2n=960?
n being the original price
\(20\%\) off means you only pay 80% of original price. So that would be .8n = 960
wow
wow such deal amazing price wow
well screw me lol
I was like yeh man that makes sense. Then like boom .8 is the original price payed
n=1200 That makes sense cause 960/.2 is like 4000 so yeh thats a lot more sense
yeah lol
and 1200-240=960
3y=2x-5 y=2/3x-5/3
correct
or y=.67x-1.67
often we prefer to just leave it as fraction
Alrights will do :p
Next one would it look like x(x-5-7)
get all term to one side So you have \(x^2-5x-14\)
Well damn I just realized I can't cut it in half cause 5 is a prime number so I would have to multiply it by 2 and that would be 2*5 and thats 10 egh
the most you can do its take x out?
well, we just factor into something like \((x-a)(x+b)\)
Oh gawd more facepalms rolling out
Cause we legit just did that like 2 problems ago ._.
well 3
So how would we deal with 5 ?
cause other than the 5 i know how to solve this
well, you can add itself by 2x, then you also subtract by 2x, to keep" balance" So you have \(x^2-5x-14+ 2x-2x~~~\Rightarrow~~~x^2+2x-7x-14\) Now can you factor?
(x+2)(x+7)
Or to factor, you can ask yourself ab = 14 b+a = -5 So what are a and b?
close. it's \((x+2)(x-7)\)
oh damn didnt realize the subtraction signs
So for the next one can we do 6x^2 +3x + 4x +6?
i mean +2
yeah
Got it !(3x+2)(2x+1)
yay lol
so to complete the square of the next problem you would do b=2 (2/2)^2 = 1 x^2 + 2x +1 = 5+1 (x+1)^2 =6
i think
i had to like triple check this lel
Actually it's \((x+1)^2=-4\)
Wow hows?
Where did I go wrong
You already have 5 there, and you want 1, so you subtract both sides by 4
Oh welp that would make sense
so as you can see. no real root.
because when you square any real number, it is always positive. But here, you have -4/
So for the next one then so you divide both sides by 2 to x^2 alone so x^2-7/2x+3=0
((7/2)/2)^2 so 3.1 is the decimal
you only need b/2 So you would have 7/4 \((x-\frac74)^2\)
The like rules they giving me is to get x^2 secluded s:

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