subtract: y/y-2 - 8+y/2-y

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subtract: y/y-2 - 8+y/2-y

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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This is tricky... you need to know this fact. 2 - y = -(y - 2) you can always flip numbers around a - sign and multiply by a - so 5 - x = -(x - 5)
In this case you want the 2 - y to be y - 2 because you need a common denominator. so you turn the bottom in to y - 2 and then you put a - sign out in front... In this case that negative sign out front would turn your problem into a + sign y/y-2 + 8 + y/y - 2 Now you have common denominators y + 8 + y -------- y - 2 2y + 8 ------ y - 2
omg hi again!!!

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this was our next question
OK... I just found you... do you get this?
i dont understand how you get the common denom....
Well... y - 2 and 2 - y are opposites of each other.... you should be able to see that if you flip the one on the right around the minus sign it becomes the y - 2 ex. 2 - y flips to y - 2 OK?
then you have to change the subtraction to addition??
Yes... because it is the opposite you have to put a - sign out in front, and actually you change the sign of the problem... so in this case the subtraction becomes addition. This is a hard concept.
hmmm then what
Then you have a common denominator... so if you look above to my steps you will see that you put everything on the top over y - 2 because you now have a common denominator
7/T-2 - 6/T^2-2T - 3/T is the LCD T^2-2T?
7 6 3 ---- - ----------- - ------ T - 2 T^2 - 2T T T - 2 factors into (T - 2) T^2 - 2T factors into T(T - 2) T factors into T Common denominator is T(T - 2) or T^2 - 2T easier if you write it T(T - 2)
T(T - 2) * 7 T(T - 2) * 6 T(T - 2) * 3 ------------ - ---------- - ---------- (T - 2) T(T - 2) T
The (T - 2)'s cancel in the first fraction The T and (T - 2) cancel in the second fraction The T cancels in the third fraction
7t-6-3t-2 ??
7T - 6 - 3(t - 2) 7T - 6 - 3T + 6 4T
isnt it -6?
left with 4t-12
The - sign goes with the 3 so it is actually -3(T - 2) and -3 * -2 = +6
ohhh got it
meeee tooooooo!
YEAH!!!!!
5-x/x-1 / 2-x/x+1
5 - x 2 - x ---- divided by ----- x - 1 x + 1 write the divide sign that has a --- with a dot above and below. Are you sure this is what you want.
noooo wait hold on
X 5- ---- X-1 ___________ X 2- ---- X+1
yes!
OK... First we will simplify the top. 5 x - - ---- 1 x- 1 what is the common denominator?
x-1?
(x - 1)*5 x ------- - ----- x - 1 x - 1
5x - 5 x ----- - ----- x - 1 x - 1 4x - 5 ------ x - 1 ok?
5X+1-X --------- X-1
GOT IT
Yes... now add the x's
Now let's do the same to the bottom.
X+2 ----- X+1
2 x - - ---- 1 x + 1 (x + 1)*2 x -------- - ---- (x + 1)*1 x + 1 (x + 1) *2 x -------- - ----- x + 1 x + 1 2x + 2 - x --------- x + 1 x + 2 ---- x + 1
OOPS... go back to the first one... you have 4x - 1 look at mine... it is 4x - 5.. you have to distribute.
got it
4x - 5 x + 2 ------ divided by ------ x - 1 x + 1 Now.... when ever you have a fraction divided by a fraction. you ALWAYS multiply by the reciprocal
4X+5 ------ X+1 -------- X+2 --- X+1
4x - 5 x + 1 ------ x ----- x - 1 x + 2
top right and bottom left cancel out?
4X+5 ------ X+2
amazing... go back and look at the numerator.. it is 4x - 5
4X-5 ------ X+2 ????? I HOPE
I wish they did cancel... they usually do.. but not here a (x - 1) can ONLY cancel with an (x - 1) and a (x + 1) can ONLY cancel with (x + 1) Usually you CAN cancel, but not in this problem. You would multiply the numerators (FOIL METHOD) then multiply the denominators (FOIL METHOD) and you would have your answer
so now what!?!?!!? omg im lost now
(4x - 5)(x + 1) ------------- (x - 1)(x + 2) 4x^2 + 4x - 5x - 5 ---------------- x^2 + 2x - x - 2 4x^2 - x - 5 ----------- x^2 + x - 2
Unless you did not type in the original problem correctly? otherwise this is the answer.
i got that
Good.
i wanna cry this is hard
solve for X: (4x-7)^2 = -27 :)
They are just a lot of steps.
(4x - 7)^2 = -27 Are you sure this is correct?
Yes
yessssssssss
OK.. so you know what imaginary numbers are i's
yes !
OK.. what if I gave you this problem... do you know what to do. x^2 = 16
yes square root both sides
Good... so in this problem you have (something)^2 = -27 so you square root both sides
x=4
(4x - 7)^2 = -27 4x - 7 = sqrt(-27)
sqrt(-27) is -3i sqrt(3) do you understand this?
\[3i \sqrt{3?}\]
yess!
OOPS you are correct is is positive
4x - 7 = 3i sqrt(3) 4x = 3i sqrt(3) + 7 x = 3i sqrt(3) + 7 ------------ 4
\[9+3x-2x ^{2} \over x ^{2}-16 \]
wait im not done hold on!
all that divided by \[4x ^{3}-9x \over 2x ^{2}+5x-12\]
simplify and state restricitons
-2x^2 + 3x + 9 2x^2 + 5x - 12 --------------- x ------------- x^2 - 16 4x^3 - 9x since it is a fraction divide by a fraction.. we are going to turn it into a multiply problem right away by using the reciprocal. OK?
ok hold on lemme take all that in
then factor?
yes we have to factor all four... but it is difficult to factor when there is a negative in front of the x^2 so for the first fraction in the numerator we are going to factor out a - sign first. - (2x^2 - 3x - 9)
-(2x + 3)(x - 3) (2x - 3)(x + 4) -------------- x ------------- (x + 4)(x - 4) x(4x^2 - 9) x(2x + 3)(2x - 3)
-(2x+3)(x-3) (x+4) (2x-3) ---------- ------------ (x+4) (x-4) x(2x+3) (2x-3)
now we can cancel out right?
Yes... now cancel anything in ( )'s with like ( )'s Anything on the top with anything on the bottom that matches.
x-3 ----- x(x-4)
close you still have that - sign on top -(x - 3) ------- x(x - 4) they may want it like -x + 3 ------ x^2 - 4x it is all the same...
Now you also said you had to tell restrictions.
yesss
OK... go back to the factored fractions. The rule is you can NEVER divide by 0 so whatever makes those ( )'s 0 would be a restriction.
We are only concerned with the denominators.
okkk
(x + 4)(x - 4) x(2x + 3)(2x - 3)
-4,4 -3/2, 3/2
And one more you need 0 because of the single x Every where there is an x there is a restriction.
blexting you are great!!! we have 3 more can we keep going?
Sure
way better then our professor!
X^2-9 ---------------- X^3+X^2-9X-9 SIMPLIFY
Factor
(x + 3)(x - 3) ------------
When you have 4 terms you should try this first. Take the first two terms and take out a common factor. x^3 + x^2 = x^2(x + 1) Take the last two terms and do the same thing. -9x - 9 = -9(x + 1) OK?
x+1 is the answer right?
(X^2-9) (X+1)
I MEAN (X-1)
OK.. Now you have x^2(x + 1) - 9(x + 1) ... good amozing. because (x + 1) is a common factor you now have (x + 1)(x^2 - 9) BUT this is the difference of two squares. (x + 1)(x + 3)(x - 3)
they dont cancel out
?
So you have the top which is (x + 3)(x - 3) ------------ (x + 1)(x + 3)(x - 3) then cancel and you are left with 1 ------- x + 1 with restrictions of x cannot be -1,-3 or 3
solve for X: 3 17 7 ------- = ---- - ----- 4x+2 2 2x+1
Common denominator 4x + 2 = 2(2x + 1) 2 = 2 2x + 1 = 2x + 1 common denominator 2(2x + 1)
Put that on top of each fraction and then cancel with the denominators.
2(2x + 1) * 3 2(2x + 1) * 17 2(2x + 1) * 7 ----------- = ------------- - ----------- 2(2x + 1) 2 2x + 1
34X+3=3 ??
X=1/34
3 = 17(2x + 1) - 14 3 = 34x + 17 - 14 3 = 34x + 3 0 = 34x 0 = x
amozing when you take 34x + 3 = 3 and you subtract 3 you get 34x = 0
Restrictions x cannot be -1/2
GOT IT, THANKS!! ONE MORE :)
OK
X^2+XY-2Y^2 X ----------- X ----------------- X^3+X^2Y X^2+3XY+2Y^2
3X^3Y^2
NOT 3Y JUST Y SORRY
(x + 2y)(x - y) x ------------ times -------------- x^2(x + y) (x + 2y)(x + y)
(x - y) ------------ x(x + y)^2
GOT IT THANKS!!
gixxergirl..... did you leave us?
Are we done????
lol sorry i had to pee!!! but thank you so mucchhhhh for everythinggg !!! are you a math teacher?
yes I am in a high school
YOU HAVE BEEN SO HELPFUL!! THANK YOU SO MUCH
No problem.. if you see me on here again and want help.. just holler. Have a wonderful Easter... He is risen!!! Good night.

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