2007-09-26 11:20:24 · 6 answers · asked by theblondebrat31 1 in Science & Mathematics ➔ Mathematics
let's times the bracket out first,
3y + 14y-28 = y-12
get the 'y's over to one side, numbers on the other
3y + 14y - y = - 12 + 28
16y = 16
y = 1
2007-09-26 11:31:22 · answer #1 · answered by yomper 3 · 0⤊ 0⤋
3y+7(2y-4)=y-12; distribute
3y + 14y - 28 = y - 12; combine terms on the left
17y - 28 = y - 12; subtract y from both sides
16y - 28 = -12; add 28 to both sides
16y = 16; divide by 16
y = 1
2007-09-26 11:25:49 · answer #2 · answered by sfroggy5 6 · 0⤊ 0⤋
3y+7(2y-4)=y-12
3y+14y-28=y-12
16y=16
y=1. ANS.
2007-09-26 11:31:18 · answer #3 · answered by Anonymous · 0⤊ 0⤋
3y + 7(2y-4) = y-12
multiply
3y + 14y - 28= y-12
combine like terms
17y - 28= y-12
16y - 28= -12
16y = 16
divide
y= 1
2007-09-26 11:25:18 · answer #4 · answered by imajesuslover21 2 · 0⤊ 0⤋
3y+14y-28=y-12 distribute
16y=16 combine like terms
y=1
2007-09-26 11:29:56 · answer #5 · answered by will 3 · 0⤊ 0⤋
3 y + 14 y - 28 = y - 12
16 y = 16
y = 1
2007-09-30 07:08:30 · answer #6 · answered by Como 7 · 0⤊ 0⤋