Combine as indicated by the signs:
4/y^2-9 + 5/y+3

- anonymous

Combine as indicated by the signs:
4/y^2-9 + 5/y+3

- schrodinger

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- UsukiDoll

do you know how to factor? and take the least common denominator?

- anonymous

...unfortunatly i dont really I'm in the process of teaching myself.

- UsukiDoll

\[\frac{4}{y^2-9} + \frac{5}{y+3}\]

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## More answers

- UsukiDoll

there is a perfect square for the first fraction in the denominator ... a product of some integer with itself. \[y^2-9\]

- UsukiDoll

in the form of \[(a^2-b^2) =(a+b)(a-b)\]
so if we let a = y^2 and b=9
we need to take the square root of y^2 and the square root of 9. Can you do that?

- anonymous

wouldnt the square root of y^2 just be y and the square root of 9 is 3?

- UsukiDoll

yay!

- UsukiDoll

so since we have a = y and b = 3 we have \[(y^2-9)=(y+3)(y-3) \]

- UsukiDoll

\[\frac{4}{(y+3)(y-3)} + \frac{5}{y+3}\]

- UsukiDoll

now noticed how our denominators are different? :O
we need to take the lcd..
since lcd is (y+3)(y-3)
we need to multiply ?????????????? so the denominator in the second fraction is the same as the denominator in the first fraction

- UsukiDoll

so what is missing in the denominator for the second fraction? I have y+3 but I don't have ?????

- anonymous

y-3? im kinda lost.

- UsukiDoll

you're on the right track

- UsukiDoll

so we multiply y-3 on the numerator for the second fraction
and multiply y-3 on the denominator for the second fraction as well

- UsukiDoll

\[\frac{4}{(y+3)(y-3)} + \frac{5}{y+3} \times \frac{y-3}{y-3}\] your fraction will look something like this

- UsukiDoll

\[\frac{4}{(y+3)(y-3)} + \frac{5(y-3)}{y+3(y-3)} \]
so we distribute the 5 all over y-3

- UsukiDoll

can you distribute the 5 towards y-3 ? it's like multiplying
5 times y and 5 times -3

- anonymous

5y-15?

- UsukiDoll

\[\frac{4}{(y+3)(y-3)} + \frac{5y-15}{y+3(y-3)}\] yes

- UsukiDoll

\[\frac{4+5y-15}{(y+3)(y-3)} \] now we need to compute 4-15

- anonymous

-11

- UsukiDoll

correct

- UsukiDoll

\[\frac{5y-11}{(y+3)(y-3)}\]

- UsukiDoll

we can't factor anything out and there's nothing to cancel, so we're done

- anonymous

ok i got it and can you look at one more problem and tell me why my answer is only partially right?

- UsukiDoll

sure

- anonymous

8-y/3y + y+2/9y - 2/6y
i got 2y+23/9y

- UsukiDoll

\[\frac{8-y}{3y}+ \frac{y+2}{9y}-\frac{2}{6y}\] this?

- anonymous

yes

- UsukiDoll

ok so all of them have y in common but we notice that the denominators are different.. we also notice that the third fraction \[\frac{2}{6}\] can be reduced . can you reduce that fraction for me?

- anonymous

1/3

- UsukiDoll

\[\frac{8-y}{3y}+ \frac{y+2}{9y}-\frac{1}{3y}\] alright... now we noticed that the first and third fraction has the same denominator, so we can rewrite this fraction and solve .
\[\frac{8-y}{3y}-\frac{1}{3y}+ \frac{y+2}{9y}\]
\[\frac{8-y-1}{3y}+\frac{y+2}{9y}\]
can you combine like terms on the first fraction ?

- UsukiDoll

we just have to do this arithmetic
8-y-1 which can be rewritten as -y+8-1

- anonymous

ok

- anonymous

?

- UsukiDoll

so what is the answer?

- UsukiDoll

I can't combine variables... but I can combine numbers for
-y+8-1

- anonymous

so would it just be -1+8-1/9y?

- UsukiDoll

no.. leave -y alone .. what is 8-1?

- anonymous

7

- UsukiDoll

\[\frac{-y+7}{3y}+\frac{y+2}{9y}\] yes.. we still can't add these denominators.. we have the y's so we just need a number
so 3 x ? = 9

- anonymous

3

- UsukiDoll

\[\frac{3(-y+7)}{9y}+\frac{y+2}{9y}\]

- UsukiDoll

now distribute the 3
what is 3 times -y
what is 3 x 7 ?

- anonymous

3 x 7=21 3 x-y=-3y

- UsukiDoll

\[\frac{(-3y+21)}{9y}+\frac{y+2}{9y}\]

- UsukiDoll

so what is -3y+y
what is 21+2

- UsukiDoll

\[\frac{(-3y+y+21+2)}{9y}\]

- anonymous

23 and -4y

- UsukiDoll

23 is correct but for the y portion that's wrong

- UsukiDoll

if you're adding a big negative number and a small positive number, your answer should go down.

- UsukiDoll

\[\frac{(-3y+y+23)}{9y}\]

- UsukiDoll

try again what is -3y+y

- UsukiDoll

hmmm... let's ignore the y part ... let's just solve -3+1
think of it as... you want to buy something for $3 but you only have $1, how much more do you need to buy that product?

- anonymous

2

- UsukiDoll

yeah, but there's one problem... you're $2 in the red in that example, so
-2 is the answer

- UsukiDoll

\[\frac{(-2y+23)}{9y}\]

- UsukiDoll

you had the correct answer but you forgot the - sign on the 2y.. I highly suggest you practice on adding and subtracting with negative numbers

- anonymous

i see now that makes more sense and i totally will work on it ha! thanks

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