add simplify if possible: 7/y-1 + 4/(y-1)^2 =
what is a common denominator equal to ?
not understanding your question
do you know the definition of numerator and denominator? what is the common denominator means?
the denominators are y-1 and y-2^2
so, to be able to add two fractions you need COMMON denominator, right? what is the common denominator in this case?
correction y-1 and y-1^2
i got that... happens sometime
y=-2
so, one fraction has denominator: (y-1) another: (y-1)^2, which is = (y-1)*(y-1) right?
break them down into 2 equeations: 7/y-1 abd 4(y-1)^2
no
ok then what this stuff is new to me
common demonotator for 7/y-1 is y. since y x 1 is 1y they cancel out so its just 7
=\[=[7(y-1) +4]/(y-1)^{2}=(7y-3)/(y-1)^{2}\]
when you have a number before the bracketsd it means you have to multiply every number by that number so it would be (4y-4)^2
so its (4y-4)(4y-4) and you should know it from there
biljana92000 ... what are you talking about...? unless I'm reading the original problem wrong. jazi1, please check the problem & let me know if this is right: \[7/(y-1) +4/(y-1)^{2}=...\] did i read it wrong?
that's correct inik but the first y-1 is not in parenthese
do you mean its 7 divided by y after that -1 or 7 divided by (y-1)?
it just like you have but the first (y-1) is not in parenthese but the second one (y-1)^2 is
again, parentheses or not - different people on web using shortcuts in writing problems & it's OK . I just trying to understand which problem you are trying to solve... it will have VERY different results 1) \[7/y +4/(y-1)^2 -1\] (I change the order just to see better the problem) 2)\[7/(y-1) +4/(y-1)^2\] which means that 7 is divided by y-1, not by1which one is your problem - 1st or 2nd?
if second - you already have the answer above if first... I'll post it
7/(y-1) + 4/(y-1)^2
so inik the answer is -2
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