Question

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Answer 1

@DarkBlueChocobo

Answer 2

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

Answer 3

Hay there

Answer 4

so then I went

Answer 5

\[\frac{ 2x-3 }{ 4 }=\frac{ 10-15x }{ 6 }\]

Answer 6

Yes, then you cross-multiply

Answer 7

\[12x-18=40-60x\]

Answer 8

\[72x-18=40\]

Answer 9

yeah

Answer 10

\[72x=58\]

Answer 11

Is that enough to tell what kind of equation it is?

Answer 12

you can tell that x is going to equal something, so it wont be indentity.

Answer 13

two other options are conditional equation and inconsistent equation

Answer 14

Inconsistant i believe?

Answer 15

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.

Answer 16

Oh so conditional is just like a usual equation that works out ?

Answer 17

If you have something like x+1 = x+2, it's inconsistent equation because you end up with 1=2

Answer 18

yeah

Answer 19

lel wow the facepalms are coming out today boys

Answer 20

1c. 0=8 so this is false or inconsistent?

Answer 21

pretty much both

Answer 22

I am having many issues on 2a

Answer 23

inconsistent equation is kind of fancy way to say false equation lol

Answer 24

you would need to combine fraction on right side.

To do so, you need to get denominator to be same

Answer 25

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

Answer 26

\[\dfrac{1}{y-2}-\dfrac{2}{y+1}\quad\longrightarrow\dfrac{y+1}{y+1}\cdot\dfrac{1}{y-2}-\dfrac{2}{y+1}\cdot\dfrac{y-2}{y-2}\]

Answer 27

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

Answer 28

kinda hard for me to explain lol

Answer 29

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

Answer 30

\[\dfrac{y+1}{y+1}\cdot\dfrac{1}{y-2}-\dfrac{2}{y+1}\cdot\dfrac{y-2}{y-2}\\~\\= \dfrac{y+1}{(y+1)(y-2)} - \dfrac{2(y-2)}{(y+1)(y-2)}\\~\\=\dfrac{y+1-2(y-2)}{(y+1)(y-2)}\]

Answer 31

ok let's start with something simply.

\(\dfrac{1}{6}+\dfrac{1}{2}\)

How can we combine fraction here?

Answer 32

multiply by 3/3

Answer 33

so 1/6+1/2*3/3

Answer 34

so you would get 1/6+3/6

Answer 35

right, so for problem 2a. it's same idea, just with variable.

Answer 36

get what I am saying here?

Answer 37

Yes I do its just weird for me to visualize and see it right away t-t idk why

Answer 38

because it's something new for you, I guess?

Answer 39

Well like the bigness of the equation kinda like put me in my place kinda deal

Answer 40

ok, so now we have \[\dfrac{y+1-2(y-2)}{(y+1)(y-2)}\]

simplify numerator

Answer 41

\[y+1-2y+4\]

Answer 42

CLEP test seem little harsh lol...

Answer 43

don't forgot combine like term

Answer 44

yeh was just showing steps

Answer 45

-y+5

Answer 46

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

Answer 47

Right.

Now cross multiply

\[\dfrac{7}{y^2-y-2} = \dfrac{-y+5}{(y-2)(y+1)}\longrightarrow~?\]

Answer 48

so we have to do the same with the other side then get (y-2)(y+1) as denominator?

Answer 49

oh i guess we dont

Answer 50

Not really. denominator doesn't matter if they are on different side of equation.

Answer 51

\[y^2(-y+5) ,-y(-y+5),-2(-y+5)\] would this work?

Answer 52

so I got -y3 + 5y^2
y^2 + 5y
2y+10

Answer 53

before combining

Answer 54

-y^3 + 5y^2
y^2 - 5y
2y-10

Answer 55

damn i messed up multiplying

Answer 56

you just messed up with +/- signs

Answer 57

\[-y^2+6y^2-3y-10\]

Answer 58

yeah

Answer 59

now the other side

Answer 60

let me delete that i need to slow down

Answer 61

7(y+1)
7(y-2)
14y+7, 14y-14
14y+14y, 7-14

Answer 62

none. left side should start with \(7(y+1)(y-2)\)

Answer 63

so just keep it as is o.o?

Answer 64

So firstly, you expand \((y+1)(y-2)\)

Answer 65

why not work out?

Answer 66

You only distribute something with plus sign.

\(a(b+c) = ab+ac\)

But \(a(bc) = abc\)

Answer 67

Alrights

Answer 68

make sense?

Answer 69

Yes it does

Answer 70

y^2+-2y+y-2?

Answer 71

Yeah

Answer 72

so we have \[y^2-y-2=-y^3+6y^2-3y-10\]

Answer 73

you forgot 7

Answer 74

\[7(y^2-y-2)=-y^3+6y^2-3y-10\]

Answer 75

damn

Answer 76

so do we expand it out with 7 now?

Answer 77

yup

Answer 78

\[7y^2-7y-14=-y^3+6y^2-3y-10\]

Answer 79

I got that

Answer 80

now take all terms to right side;

\(0 = -y^3-y^2+4y-4\)

Answer 81

can confirm got that after working out

Answer 82

lel sorry I doing this in a note book :p

Answer 83

know what to do from here?

Answer 84

Notice something you can factor out?

Hint: \[0 = \boxed{-y^3-y^2}+\boxed{4y-4}\]

Answer 85

we can simplify ?

Answer 86

oh you asked lol

Answer 87

Oops should be \(0 = -y^3-y^2+4y+4\)

Answer 88

would we simplify the y's?

Answer 89

well, actually we try to factor, so we can easily figure out solutions.

Answer 90

-1(y^2+4)(y+1)

Answer 91

Would that be how we factor it?

Answer 92

close. should be \((y+1)(4-y^2)\)

Answer 93

Then you can factor \(4-y^2\)

Answer 94

or you can "factor out" -1 from \(4-y^2\)

So you would have something like \((-1)(y^2-4)\)

Answer 95

-y^2+4. Doesnt this just put us back where we were?

Answer 96

Yeah, just to make it slightly easier to factor out, but never mind.

Answer 97

factoring -1 out is just optional.

Answer 98

You got \(0 = (y+1)(2-y)(y+2)\)?