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The question isn't complete
Please provide the instructions that came with this expression. What's your goal? Can you factor one or both of the denominators?
srry....."Combine as indicated by the signs."
make the denominators same
how would i do that??
factor both denominators if possible. Determine the lowest common denominator (LCD). Use the LCD to combine these fractions.
Hint: factor the denominator of the first fraction first. Do the factors have anything in common with the denominator of the second fraction?
yes. 'y' and 3*3 = 9
\[y^2-9=y^2-3^2\] is the "difference of two squares." Can y ou factor this? look at y^2 - 3^2 as a whole and come up with 2 factors.
\[a^2-b^2=(a-b)(a+b)\]
sooooo (y-3)(y+3)??
Very good. Compare these two factors to the den. of the 2nd fraction. Describe your observations.
cancel out the y+3
Can't. The idea here is to change the 2nd fraction so that its denom. is the same as that of the first fraction.
How would you do that? \ On the left you have the den. \[(y-3)(y+3). \]
On the right you have the denom. y+3. What's missing there? What are you going to do about it?
the right already has the y+3 so all i need to get is the y-3
lol.......ummm.... -1?
Yes. the 2nd fraction is \[\frac{ 5 }{y+3}\]
Put that y+3 inside parentheses. Next, multiply numerator (5) and den. (y+3) by (y-3). Results?
if i did it right...... 5/9y? or 5/ y+9
The second fraction is \[\frac{ 5 }{ y+3 }.\]
I was asking you to multiply the numerator (5) by (y-3), and also to mult. the denom. (y+3) by (y-3). Please do that now. Results?
ohh srry........ 5y-3/y+9 srry im horrible at math
Here's what you've typed in: 5y-3/y+9 Here's what we should have: 5(y-3) ----------- (y+3)(y-3) Your numerator would be better if you'd enclose that (y-3) inside parentheses.
Be certain you understand what I've done here; this is important. The goal was to make the denominators of the 2 fractions the same.
We now have \[\frac{ 4 }{( y-3)(y+3) }+\frac{ 5(y-3) }{ (y+3)(y-3) }\]
I know they have to b the same, it getting them the same thats the problem for me..... but i understand. I kinda remember going ove the perfect squares and factoring with my teacher
Are the denominators now the same?
now I just work that out right?
switched around........?
That switch of order doesn't matter here. The dens. really are the same, mathematically. Can you multiply out 5(y-3)?
would it b 5y-15?
yes, it would
and therefore, our \[\frac{ 4 }{( y-3)(y+3) }+\frac{ 5(y-3) }{ (y+3)(y-3) }\]
becomes\[\frac{ 4 }{( y-3)(y+3) }+\frac{ 5y-15 }{ (y+3)(y-3) }\]
then i add em right?
Can you now add these 2 fractions together? Yes, you do "add em."
so.... ummmmm 5y-11/...... (y-6)(y+6)?
Look at the numerators: You have 4 + 5y -15. Yes, you have added these correctly. But why have you changed (y-3)(y+3) to (y-6)(y+6)?
so i dont add those of course they r denominators. srry wasnt thinking those stay the same right?
the denominators stay exactly the same. Note that \[(y-3)(y+3)=y^2-3^2=y^2-9\]
Please write your final result here. No need to multiply out the denoms.
5y-11/ (y-3)(y+3) or 5y-11/y^2 -9
the bottom one right?
It would be right if you'd please enclose the 5y-11 inside parentheses.
(5y-11)/ y^2 -9
Great. This is the answer you wanted. You have combined the 2 fractions. any questions?
I understand it more than I did earlier thanks!!
Please practice this. See whether you can copy down the given problem and combine the 2 frac tions into one on your own. Hope to work with you again soon.
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