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Mathematics 8 Online
iosangel:

Mth help

iosangel:

1 attachment
iosangel:

@vocaloid

iosangel:

I tried 2x then +3

iosangel:

It wast ritgh, but i dont understand how else???

iosangel:

wasnt*

XLynx:

this stuff is child's play man what math class are you in?

iosangel:

?

iosangel:

review for precalc

XLynx:

what math are you taking in school

XLynx:

same, what grade?

iosangel:

are you going to help?

XLynx:

yeah, one min

XLynx:

2x and then 3 at the top (x+3) in the bottom first box (x-5) in the second

XLynx:

did that at the top of my head 😎

iosangel:

thats wrong

iosangel:

That is what i said in the beginning

XLynx:

those two literally sum together wth, show me how you're entering it

XLynx:

screenshot it

iosangel:

no, ive tried that with a question before i know what you're trying to do its wrong

XLynx:

thats literally how math works lol. 3/5 + 1/5 = 4/5. whats common, a common denominator

XLynx:

the top two are added together

iosangel:

Thats not right

XLynx:

one sec, ill show

1 attachment
XLynx:

see? those two equations add together to form the top one

XLynx:

(x-5)(x+3) = x^2-2x-15

iosangel:

yes ive tried that its wrong

iosangel:

i had a question similar and got it wrong

XLynx:

then the question is probably broken broski lol, or theres something in context that im not getting

XLynx:

is this from KA?

iosangel:

?

XLynx:

khan academy

iosangel:

go to msgs pls

XLynx:

sure

XLynx:

i got a 780 on SAT math so ive already graduated from this stuff lol, idk why its wrong

iosangel:

ive taken the act too, doesnt mean anything..?

XLynx:

what did you get on the ACT

XLynx:

converts to an SAT score lol

iosangel:

go to msgs so somone else can help idk how to use msgs

XLynx:

its in simplest form too which is weird

XLynx:

are you sure you're not missing anything

Vocaloid:

the expression can be simplified beyond that. let A and B be constants such that (2x+3)/(x^2-2x-15) = A/(x-5) + B/(x+3) multiply both sides by (x-5)(x+3) (2x+3) = A(x+3) + B(x-5) let x = -3 to eliminate the term with A and solve for B; likewise let x = 5 to solve for A. A and B should both be fractions which you can enter in the corresponding numerator and denominator blanks.

iosangel:

What do you mean "multiply both sides by (x-5)(x+3)"

iosangel:

both the demoninators ?

iosangel:

ok nvm i got it, thank you

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