ALgebra help please!!
2007-11-11 06:32:10 · 10 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
4y = 15 + 9
4y = 24
y = 6
2007-11-15 05:03:13 · answer #1 · answered by Como 7 · 1⤊ 0⤋
Add nine to both sides and you have 4Y = 24. Divide both sides by 4 and get 1Y = 6
Check: Y = 6
(4 X 6) - 9 = 15
24 - 9 = 15
15 = 15
2007-11-11 06:44:41 · answer #2 · answered by gzlakewood@sbcglobal.net 4 · 0⤊ 0⤋
rearange the sum so 9 + 15 = 4y. then do 15 + 9 which makes 24 then divide by 4 =y= 6
2007-11-11 06:42:00 · answer #3 · answered by quentin a 1 · 0⤊ 0⤋
4y-9=15
+9 +9
4y=24
4 4
y=6
2007-11-11 06:43:49 · answer #4 · answered by shina_ma_booboo 2 · 0⤊ 0⤋
okay so if you're solving for y
4y-9=15
+9 +9
(do the opposite: add 9 to each side, so the nine cancels out and 15+9=24)
So:
4y=24
4y/4=24/4
(4y divided by 4= y {the 4 cancels out} so you're left with y on the left. Then 24/4 {24 divided by 4} equals 6)
So:
y=24/4
y=6
TA-DA!!!!!
I hope i explained it so that it makes sense. lol
Best wishes.
Ash.
2007-11-11 06:48:02 · answer #5 · answered by Truth be told. 2 · 0⤊ 1⤋
y=6
2007-11-11 06:40:25 · answer #6 · answered by Anonymous · 0⤊ 0⤋
put all the variables on one side and all the real numbers on the other side of the equation
4y= 15 + 9
4y= 24
y= 24/4
2007-11-11 06:36:17 · answer #7 · answered by Anonymous · 0⤊ 0⤋
4y-9=15
you want to get the y isolated onto one side so:
add the 9 to the other side=> 4y=24
then divide both sides by 4
y=6
:)
2007-11-11 06:40:00 · answer #8 · answered by dudeitshanna 2 · 0⤊ 0⤋
4y-9=15
4y=24
y=6
2007-11-11 06:39:43 · answer #9 · answered by Anonymous · 0⤊ 0⤋
First, you draw a vertical line straight down the equals sign. Then, you add nine to nine, which cancels it out. so cross it out. Then, what you do to one side you gotta do to the other. So add nine to fifteen. Next, you have to divide 4 by four. That cancels the four out. So then divide 24 by 4 because what you do to one side out gotta do to the other. So Y ends up equaling 6
2007-11-11 06:42:37 · answer #10 · answered by hottie 2 · 0⤊ 0⤋