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

Hay there

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?

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

kinda hard for me to explain lol

so 1/(y+1)(y-2) - 4/(y+1)(y-2) ?

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

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?

you are confused on how to get solution?

But that wouldn't work so I just am straight confused where to go from here

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

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

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 \(x^2-5x-14\)

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

(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

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: