How would you figure out a problem like this...I'm mainly concerned on the steps. Thanks!
[(3y-2)/6] - [(y-3)/9]
2007-03-08 03:22:12 · 4 answers · asked by kcdude 5 in Science & Mathematics ➔ Mathematics
(3y - 2)/6 - (y - 3)/9
Your first goal is to give them a common denominator; find the lowest common denominator of 6 and 9; it's 18. Make each fraction over 18.
3(3y - 2)/18 - 2(y - 3)/18
Now, merge these two fractions into one.
[3(3y - 2) - 2(y - 3)]/18
Simplify the top.
[9y - 6 - 2y + 6]/18
Group like terms.
7y/18
2007-03-08 03:28:01 · answer #1 · answered by Puggy 7 · 0⤊ 0⤋
(3y-2)/6-(y-3)/9
[First of all we find out the LCD or LCM of 6 &9
which is 18.]
={3(3y-2)-2(y-3)}/18 [the two fraction is added]
=(9y-6-2y+6)/18 [numerator is being simplified]
=7y/18 [simplification of numerator completed]
2007-03-08 11:39:33 · answer #2 · answered by alpha 7 · 0⤊ 0⤋
.[(3y-2)/6] - [(y-3)/9]
1.find lowest common denominator of 6 and 9.its 18.6 goes into 18 3 times and 9 goes into 18 2 times.
therefore .[(3y-2)/6] =3(3y-2)/18=(9y-6)/18
and [(y-3)/9]=2(y-3)/18=(2y-6)/18
so .[(3y-2)/6] - [(y-3)/9]=(9y-6)/18 -(2y-6)/18=(9y-6-2y+6)/18=7y/18
2007-03-08 11:27:35 · answer #3 · answered by hiphop 2 · 0⤊ 0⤋
= y / 2 - 1/3 - y / 9 + 1/3
= 9y /18 - 2y /18
= 7y / 18
2007-03-08 11:54:15 · answer #4 · answered by Como 7 · 0⤊ 0⤋