geerky42
  • geerky42
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Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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geerky42
  • geerky42
@DarkBlueChocobo
geerky42
  • geerky42
\(\dfrac{2x}{4}-\dfrac{3}{4}\quad\Longleftrightarrow\dfrac{2x-3}{4}\)
DarkBlueChocobo
  • DarkBlueChocobo
Hay there

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DarkBlueChocobo
  • DarkBlueChocobo
so then I went
DarkBlueChocobo
  • DarkBlueChocobo
\[\frac{ 2x-3 }{ 4 }=\frac{ 10-15x }{ 6 }\]
geerky42
  • geerky42
Yes, then you cross-multiply
DarkBlueChocobo
  • DarkBlueChocobo
\[12x-18=40-60x\]
DarkBlueChocobo
  • DarkBlueChocobo
\[72x-18=40\]
geerky42
  • geerky42
yeah
DarkBlueChocobo
  • DarkBlueChocobo
\[72x=58\]
geerky42
  • geerky42
Is that enough to tell what kind of equation it is?
geerky42
  • geerky42
you can tell that x is going to equal something, so it wont be indentity.
geerky42
  • geerky42
two other options are conditional equation and inconsistent equation
DarkBlueChocobo
  • DarkBlueChocobo
Inconsistant i believe?
geerky42
  • geerky42
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.
DarkBlueChocobo
  • DarkBlueChocobo
Oh so conditional is just like a usual equation that works out ?
geerky42
  • geerky42
If you have something like x+1 = x+2, it's inconsistent equation because you end up with 1=2
geerky42
  • geerky42
yeah
DarkBlueChocobo
  • DarkBlueChocobo
lel wow the facepalms are coming out today boys
DarkBlueChocobo
  • DarkBlueChocobo
1c. 0=8 so this is false or inconsistent?
geerky42
  • geerky42
pretty much both
DarkBlueChocobo
  • DarkBlueChocobo
I am having many issues on 2a
geerky42
  • geerky42
inconsistent equation is kind of fancy way to say false equation lol
geerky42
  • geerky42
you would need to combine fraction on right side. To do so, you need to get denominator to be same
geerky42
  • geerky42
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
geerky42
  • geerky42
\[\dfrac{1}{y-2}-\dfrac{2}{y+1}\quad\longrightarrow\dfrac{y+1}{y+1}\cdot\dfrac{1}{y-2}-\dfrac{2}{y+1}\cdot\dfrac{y-2}{y-2}\]
geerky42
  • geerky42
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
geerky42
  • geerky42
kinda hard for me to explain lol
DarkBlueChocobo
  • DarkBlueChocobo
so 1/(y+1)(y-2) - 4/(y+1)(y-2) ?
geerky42
  • geerky42
\[\dfrac{y+1}{y+1}\cdot\dfrac{1}{y-2}-\dfrac{2}{y+1}\cdot\dfrac{y-2}{y-2}\\~\\= \dfrac{y+1}{(y+1)(y-2)} - \dfrac{2(y-2)}{(y+1)(y-2)}\\~\\=\dfrac{y+1-2(y-2)}{(y+1)(y-2)}\]
geerky42
  • geerky42
ok let's start with something simply. \(\dfrac{1}{6}+\dfrac{1}{2}\) How can we combine fraction here?
DarkBlueChocobo
  • DarkBlueChocobo
multiply by 3/3
DarkBlueChocobo
  • DarkBlueChocobo
so 1/6+1/2*3/3
DarkBlueChocobo
  • DarkBlueChocobo
so you would get 1/6+3/6
geerky42
  • geerky42
right, so for problem 2a. it's same idea, just with variable.
geerky42
  • geerky42
get what I am saying here?
DarkBlueChocobo
  • DarkBlueChocobo
Yes I do its just weird for me to visualize and see it right away t-t idk why
geerky42
  • geerky42
because it's something new for you, I guess?
DarkBlueChocobo
  • DarkBlueChocobo
Well like the bigness of the equation kinda like put me in my place kinda deal
geerky42
  • geerky42
ok, so now we have \[\dfrac{y+1-2(y-2)}{(y+1)(y-2)}\] simplify numerator
DarkBlueChocobo
  • DarkBlueChocobo
\[y+1-2y+4\]
geerky42
  • geerky42
CLEP test seem little harsh lol...
geerky42
  • geerky42
don't forgot combine like term
DarkBlueChocobo
  • DarkBlueChocobo
yeh was just showing steps
DarkBlueChocobo
  • DarkBlueChocobo
-y+5
DarkBlueChocobo
  • DarkBlueChocobo
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
geerky42
  • geerky42
Right. Now cross multiply \[\dfrac{7}{y^2-y-2} = \dfrac{-y+5}{(y-2)(y+1)}\longrightarrow~?\]
DarkBlueChocobo
  • DarkBlueChocobo
so we have to do the same with the other side then get (y-2)(y+1) as denominator?
DarkBlueChocobo
  • DarkBlueChocobo
oh i guess we dont
geerky42
  • geerky42
Not really. denominator doesn't matter if they are on different side of equation.
DarkBlueChocobo
  • DarkBlueChocobo
\[y^2(-y+5) ,-y(-y+5),-2(-y+5)\] would this work?
DarkBlueChocobo
  • DarkBlueChocobo
so I got -y3 + 5y^2 y^2 + 5y 2y+10
DarkBlueChocobo
  • DarkBlueChocobo
before combining
geerky42
  • geerky42
-y^3 + 5y^2 y^2 - 5y 2y-10
DarkBlueChocobo
  • DarkBlueChocobo
damn i messed up multiplying
geerky42
  • geerky42
you just messed up with +/- signs
DarkBlueChocobo
  • DarkBlueChocobo
\[-y^2+6y^2-3y-10\]
geerky42
  • geerky42
yeah
DarkBlueChocobo
  • DarkBlueChocobo
now the other side
DarkBlueChocobo
  • DarkBlueChocobo
let me delete that i need to slow down
DarkBlueChocobo
  • DarkBlueChocobo
7(y+1) 7(y-2) 14y+7, 14y-14 14y+14y, 7-14
geerky42
  • geerky42
none. left side should start with \(7(y+1)(y-2)\)
DarkBlueChocobo
  • DarkBlueChocobo
so just keep it as is o.o?
geerky42
  • geerky42
So firstly, you expand \((y+1)(y-2)\)
DarkBlueChocobo
  • DarkBlueChocobo
why not work out?
geerky42
  • geerky42
You only distribute something with plus sign. \(a(b+c) = ab+ac\) But \(a(bc) = abc\)
DarkBlueChocobo
  • DarkBlueChocobo
Alrights
geerky42
  • geerky42
make sense?
DarkBlueChocobo
  • DarkBlueChocobo
Yes it does
DarkBlueChocobo
  • DarkBlueChocobo
y^2+-2y+y-2?
geerky42
  • geerky42
Yeah
DarkBlueChocobo
  • DarkBlueChocobo
so we have \[y^2-y-2=-y^3+6y^2-3y-10\]
geerky42
  • geerky42
you forgot 7
geerky42
  • geerky42
\[7(y^2-y-2)=-y^3+6y^2-3y-10\]
DarkBlueChocobo
  • DarkBlueChocobo
damn
DarkBlueChocobo
  • DarkBlueChocobo
so do we expand it out with 7 now?
geerky42
  • geerky42
yup
geerky42
  • geerky42
\[7y^2-7y-14=-y^3+6y^2-3y-10\]
DarkBlueChocobo
  • DarkBlueChocobo
I got that
geerky42
  • geerky42
now take all terms to right side; \(0 = -y^3-y^2+4y-4\)
DarkBlueChocobo
  • DarkBlueChocobo
can confirm got that after working out
DarkBlueChocobo
  • DarkBlueChocobo
lel sorry I doing this in a note book :p
geerky42
  • geerky42
know what to do from here?
geerky42
  • geerky42
Notice something you can factor out? Hint: \[0 = \boxed{-y^3-y^2}+\boxed{4y-4}\]
DarkBlueChocobo
  • DarkBlueChocobo
we can simplify ?
DarkBlueChocobo
  • DarkBlueChocobo
oh you asked lol
geerky42
  • geerky42
Oops should be \(0 = -y^3-y^2+4y+4\)
DarkBlueChocobo
  • DarkBlueChocobo
would we simplify the y's?
geerky42
  • geerky42
well, actually we try to factor, so we can easily figure out solutions.
DarkBlueChocobo
  • DarkBlueChocobo
-1(y^2+4)(y+1)
DarkBlueChocobo
  • DarkBlueChocobo
Would that be how we factor it?
geerky42
  • geerky42
close. should be \((y+1)(4-y^2)\)
geerky42
  • geerky42
Then you can factor \(4-y^2\)
geerky42
  • geerky42
or you can "factor out" -1 from \(4-y^2\) So you would have something like \((-1)(y^2-4)\)
DarkBlueChocobo
  • DarkBlueChocobo
-y^2+4. Doesnt this just put us back where we were?
geerky42
  • geerky42
Yeah, just to make it slightly easier to factor out, but never mind.
geerky42
  • geerky42
factoring -1 out is just optional.
geerky42
  • geerky42
You got \(0 = (y+1)(2-y)(y+2)\)?
DarkBlueChocobo
  • DarkBlueChocobo
Yes that would be it fully factored
geerky42
  • geerky42
yeah, now can you find solutions?
geerky42
  • geerky42
I was thinking, maybe we need to do something more efficient with time. we spend a lot of time on few problems lol.
DarkBlueChocobo
  • DarkBlueChocobo
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
geerky42
  • geerky42
you are confused on how to get solution?
DarkBlueChocobo
  • DarkBlueChocobo
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
DarkBlueChocobo
  • DarkBlueChocobo
But that wouldn't work so I just am straight confused where to go from here
geerky42
  • geerky42
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
geerky42
  • geerky42
So we can "split" equation into \(0 = y+1\) or \(0=2-y\) or \(0=y+2\)
geerky42
  • geerky42
So from here, just solve for y from each equation Does that make sense?
geerky42
  • geerky42
Who gave you medal? lol
DarkBlueChocobo
  • DarkBlueChocobo
Lol I have no clue, and Yeah that makes sense in order to get it fully factored when you have many parts
DarkBlueChocobo
  • DarkBlueChocobo
y=-1 or y=2 or y=-2
geerky42
  • geerky42
hmm Now plug y = -1 or y = 2 back into original equation
geerky42
  • geerky42
what happens?
DarkBlueChocobo
  • DarkBlueChocobo
i got 7/0=1/0-2/3
geerky42
  • geerky42
yeah. zero in denominator. So we are left with y = -2 That it lol
DarkBlueChocobo
  • DarkBlueChocobo
wait how we left with -2?
DarkBlueChocobo
  • DarkBlueChocobo
s: we still had -2/3 left ove
geerky42
  • geerky42
y = -1 and y = 2 get us zero in denominator, so they are invalid solutions.
DarkBlueChocobo
  • DarkBlueChocobo
oh that makes sense nvm so we would try y=-2
DarkBlueChocobo
  • DarkBlueChocobo
and baddabing there it comes out
geerky42
  • geerky42
You can. and you will get true equation.
DarkBlueChocobo
  • DarkBlueChocobo
Yeh Imma try that second one alone and check it laters just to make sure I understand this
geerky42
  • geerky42
alright
geerky42
  • geerky42
|dw:1434844521381:dw|
geerky42
  • geerky42
ehh not right reaction, but whatever
DarkBlueChocobo
  • DarkBlueChocobo
loool wowsers
DarkBlueChocobo
  • DarkBlueChocobo
Alright so with the first word problem., Ralph = 6-2n? Ralph + Tommy= 30?
geerky42
  • geerky42
R = 2T - 6 R+T = 30
DarkBlueChocobo
  • DarkBlueChocobo
Could we set the equation to 2t-6=30? Cause we know the total it could be
geerky42
  • geerky42
|dw:1434844843534:dw|
geerky42
  • geerky42
yeah you could do that
DarkBlueChocobo
  • DarkBlueChocobo
|dw:1434844913540:dw|
DarkBlueChocobo
  • DarkBlueChocobo
you have beautiful mouse control lol
geerky42
  • geerky42
thanks lol
DarkBlueChocobo
  • DarkBlueChocobo
3t=36 36/3 t=12
DarkBlueChocobo
  • DarkBlueChocobo
so we know Tommy scored 12 12*2 R=24-6=18
geerky42
  • geerky42
|dw:1434845125612:dw|
DarkBlueChocobo
  • DarkBlueChocobo
lol Loki pls
DarkBlueChocobo
  • DarkBlueChocobo
So this one is worded weirdly for me it sounds like we are using multiple variables?
DarkBlueChocobo
  • DarkBlueChocobo
OH wait no
DarkBlueChocobo
  • DarkBlueChocobo
twice the 3rd is 86
DarkBlueChocobo
  • DarkBlueChocobo
86/2
geerky42
  • geerky42
no
geerky42
  • geerky42
let's go ahead and set up equation
DarkBlueChocobo
  • DarkBlueChocobo
Alright
geerky42
  • geerky42
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
geerky42
  • geerky42
"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?
DarkBlueChocobo
  • DarkBlueChocobo
Yep. As I fail reading it. It sounded really weird to me egh.
geerky42
  • geerky42
yeah weird worded, I had to read twice lol.
DarkBlueChocobo
  • DarkBlueChocobo
a+4a+8+2a+8=86 7a+16=86 7a=70 a=10 First Kid=10 Second Kid= 12 Third Kid = 14
geerky42
  • geerky42
Yep!
DarkBlueChocobo
  • DarkBlueChocobo
The next one is 20%*n=960?
DarkBlueChocobo
  • DarkBlueChocobo
or .2n=960?
DarkBlueChocobo
  • DarkBlueChocobo
n being the original price
geerky42
  • geerky42
\(20\%\) off means you only pay 80% of original price. So that would be .8n = 960
DarkBlueChocobo
  • DarkBlueChocobo
wow
geerky42
  • geerky42
wow such deal amazing price wow
DarkBlueChocobo
  • DarkBlueChocobo
well screw me lol
DarkBlueChocobo
  • DarkBlueChocobo
I was like yeh man that makes sense. Then like boom .8 is the original price payed
DarkBlueChocobo
  • DarkBlueChocobo
n=1200 That makes sense cause 960/.2 is like 4000 so yeh thats a lot more sense
geerky42
  • geerky42
yeah lol
DarkBlueChocobo
  • DarkBlueChocobo
and 1200-240=960
DarkBlueChocobo
  • DarkBlueChocobo
3y=2x-5 y=2/3x-5/3
geerky42
  • geerky42
correct
DarkBlueChocobo
  • DarkBlueChocobo
or y=.67x-1.67
geerky42
  • geerky42
often we prefer to just leave it as fraction
DarkBlueChocobo
  • DarkBlueChocobo
Alrights will do :p
DarkBlueChocobo
  • DarkBlueChocobo
Next one would it look like x(x-5-7)
geerky42
  • geerky42
get all term to one side So you have \(x^2-5x-14\)
DarkBlueChocobo
  • DarkBlueChocobo
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
DarkBlueChocobo
  • DarkBlueChocobo
the most you can do its take x out?
geerky42
  • geerky42
well, we just factor into something like \((x-a)(x+b)\)
DarkBlueChocobo
  • DarkBlueChocobo
Oh gawd more facepalms rolling out
DarkBlueChocobo
  • DarkBlueChocobo
Cause we legit just did that like 2 problems ago ._.
DarkBlueChocobo
  • DarkBlueChocobo
well 3
DarkBlueChocobo
  • DarkBlueChocobo
So how would we deal with 5 ?
DarkBlueChocobo
  • DarkBlueChocobo
cause other than the 5 i know how to solve this
geerky42
  • geerky42
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?
DarkBlueChocobo
  • DarkBlueChocobo
(x+2)(x+7)
geerky42
  • geerky42
Or to factor, you can ask yourself ab = 14 b+a = -5 So what are a and b?
geerky42
  • geerky42
close. it's \((x+2)(x-7)\)
DarkBlueChocobo
  • DarkBlueChocobo
oh damn didnt realize the subtraction signs
DarkBlueChocobo
  • DarkBlueChocobo
So for the next one can we do 6x^2 +3x + 4x +6?
DarkBlueChocobo
  • DarkBlueChocobo
i mean +2
geerky42
  • geerky42
yeah
DarkBlueChocobo
  • DarkBlueChocobo
Got it !(3x+2)(2x+1)
geerky42
  • geerky42
yay lol
DarkBlueChocobo
  • DarkBlueChocobo
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
DarkBlueChocobo
  • DarkBlueChocobo
i think
DarkBlueChocobo
  • DarkBlueChocobo
i had to like triple check this lel
geerky42
  • geerky42
Actually it's \((x+1)^2=-4\)
DarkBlueChocobo
  • DarkBlueChocobo
Wow hows?
DarkBlueChocobo
  • DarkBlueChocobo
Where did I go wrong
geerky42
  • geerky42
You already have 5 there, and you want 1, so you subtract both sides by 4
DarkBlueChocobo
  • DarkBlueChocobo
Oh welp that would make sense
geerky42
  • geerky42
so as you can see. no real root.
geerky42
  • geerky42
because when you square any real number, it is always positive. But here, you have -4/
DarkBlueChocobo
  • DarkBlueChocobo
So for the next one then so you divide both sides by 2 to x^2 alone so x^2-7/2x+3=0
DarkBlueChocobo
  • DarkBlueChocobo
((7/2)/2)^2 so 3.1 is the decimal
geerky42
  • geerky42
you only need b/2 So you would have 7/4 \((x-\frac74)^2\)
DarkBlueChocobo
  • DarkBlueChocobo
The like rules they giving me is to get x^2 secluded s:

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