A few days ago
Algebra: Subtract?
These are in fraction form
g/u+g – g/u-g
Top 4 Answers
A few days ago
Favorite Answer
ok, here we go (please note: ….. is used for spacing only)
__g__ -__g__
…u+g……u-g
you need to even them up by finding the common denominator. and easy way is to multiply each by the other’s denominator.
__g__ * (u-g) – __g__*(u+g)
….u+g….(u-g)….u-g….(u+g)
which gives you:
_gu -g^2_ – _gu+g^2_ …….(the ^2 is to the 2nd power)
(u+g)(u-g)….(u+g)(u-g)
then you can bring them together:
_gu -g^2 -gu+g^2_
…..(u+g)(u-g)
this becomes
..__0__
(u+g)(u-g)……….which is 0
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A few days ago
I’m pretty sure that the answer is 0 because everything cancels each out. g/u-g/u=0 and +g_g equals zero
0
A few days ago
g + g – g -g
_ _
u u
= common denominator u
= (g+gu-g-gu)/u= 0/u= 0
0
A few days ago
take out g((1/u+1)-(1/u-1))
= g((1+u-1+u)/u)
=2gu/u
=2g
correct if theres anything u find mistake
0
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