Please help solve this equation. y^2-16/2y+6 / y-4/y+3?
回答 (6)
2009-02-11 8:56 am
✔ 最佳答案
y^2-16----------
2y+6
-----------
y-4
-------
y+3
..(y+4)(y-4).........y+3
=----------------- x ----------...y-4 and y+3 is canceled
...2(y+3)..............y-4
....y+4
=---------- answer//
.....2
2009-02-11 9:36 am
(y^2 - 16)/(2y + 6) ÷ (y - 4)/(y + 3)
= (y^2 - 4^2)/[2(y + 3)] * (y + 3)/(y - 4)
= [(y + 4)(y - 4)]/[2(y + 3)] * (y + 3)/(y - 4)
= (y + 4)/2 * 1
= (y + 4)/2
= y/2 + 4/2
= y/2 + 2
= (y^2 - 4^2)/[2(y + 3)] * (y + 3)/(y - 4)
= [(y + 4)(y - 4)]/[2(y + 3)] * (y + 3)/(y - 4)
= (y + 4)/2 * 1
= (y + 4)/2
= y/2 + 4/2
= y/2 + 2
2009-02-11 9:18 am
The answer above which is
y+4
------ is correct
2
more explanation on how to solve is to remember the rules of division and multiplication among fractions.
used here multiplication instead of division between 2 sets of fraction with equation and apply the factoring of equation to cancel both the same terms and the remaining numbers are the answer.
y+4
------ is correct
2
more explanation on how to solve is to remember the rules of division and multiplication among fractions.
used here multiplication instead of division between 2 sets of fraction with equation and apply the factoring of equation to cancel both the same terms and the remaining numbers are the answer.
參考: secret-detective.com
2009-02-11 8:54 am
(y^2 - 16/2y + 6) / (y - 4/y + 3) =
(y^2 - 16/2y + 6) * (y + 3/y - 4) =
((y + 4)(y - 4)/2(y + 3)) * (y + 3/y - 4) =
(y + 4)/2 = (y/2) + 2.
Hope this helps.
(y^2 - 16/2y + 6) * (y + 3/y - 4) =
((y + 4)(y - 4)/2(y + 3)) * (y + 3/y - 4) =
(y + 4)/2 = (y/2) + 2.
Hope this helps.
2009-02-11 8:44 am
is it each divided the next or is it two sets (y^12-16/2y+6) / (y-4/y+3)
2009-02-11 8:44 am
Hey where's "=" in th equation ?!!
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