Factor the numerators and denominators of these fractions,and then reduce the
fractions to the lowest terms
8a+8b over 12a+12b
2007-03-29 04:56:50 · 6 answers · asked by balu_srd 1 in Science & Mathematics ➔ Mathematics
8a+8b can be factored by removing the 8. It is 8(a+b).
Same with 12a+12b. It is 12(a+b).
8(a+b)/12(a+b) have the (a+b) cancelled out (whatever their values, they are identical so cancel them out!
That leaves 8/12 which can be reduced to 2/3.
Answer: 2/3
2007-03-29 05:02:41 · answer #1 · answered by kathyw 7 · 0⤊ 0⤋
2/3
2007-03-29 05:05:24 · answer #2 · answered by gebobs 6 · 0⤊ 0⤋
=8(a+b)/12(a+b)= 8/12 = 2/3
2007-03-29 05:02:42 · answer #3 · answered by ironduke8159 7 · 0⤊ 0⤋
Its 2/3 as pointed out above for all values of a except a = -b
If a = -b, then the fraction is indeterminate and you cannot cancel the (a+b) terms with meaning since they equal zero.
2007-03-29 05:08:57 · answer #4 · answered by RTB 1 · 0⤊ 0⤋
(8a + 8b)/(12a + 12b)
[8(a + b)]/[12(a + b)]
8/12
2/3
2007-03-29 05:01:20 · answer #5 · answered by JOhn M 5 · 0⤊ 0⤋
(8a+8b)/(12a+12b)=4(2a+2b)/4(3a+3b)=(2a+2b)/(3a+3b)=2(a+b)/3(a+b) =2/3
2007-03-29 05:08:25 · answer #6 · answered by llcold 2 · 0⤊ 0⤋