3y-6/5y-20*10y-40/27y-54 can be written as?

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2007-05-17 14:44:26 · 2 answers · asked by ohoma 1 in Science & Mathematics ➔ Mathematics

(3(y-2)/(27(y-2)) * (2(5y-20)/(5y-20))

(y-2) and (5y-20) cancel out

((3)/(27))*(2) = 6/27 = 2/9
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2007-05-17 14:51:20 · answer #1 · answered by Robert L 7 · 0⤊ 0⤋

(3y-6)/(5y-20) * (10y-40)/ (27y-54)
= 3(y-2) / 5 (y-4) * 10(y-4) / 27( y-2)
= 3/5 * 10/27
=2/9

2007-05-17 14:50:38 · answer #2 · answered by Ferra 2 · 0⤊ 0⤋