So I have 4 questions that I would really be glad if people helped me... im not trying to cheat or anything but ive been doing this 200 question packet over different things and I need help with these 4...
trying to find sums and differences...
(b+1/b+2) - (b+3/b+4)
(y/y^2-49) - (7/y+7)
(1/2) + (3/x+4)
(a+2/4b) - (a-1/10b) + (a-3/5b)
i know it's alot so I get if you dont answer... i just am really really confused.
2007-06-27 10:28:42 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
THE / means OVER like as in fractions
2007-06-27 10:29:35 · update #1
(b+1/b+2) - (b+3/b+4)
=[(b+1)(b+4) - (b+3)(b+2)]/ (b+2)(b+4)
=[(b^2+5b+4) - (b^2+5b+6) / (b^2+6b+8)
=-2 / (b^2+6b+8)
(y/y^2-49) - (7/y+7)
=(y-7(y-7))/(y^2-49)
=(y-7y+49/(y^2-49)
=(-6y+49)/(y^2-49)
(1/2) + (3/x+4)
=[(x+4)+6] / (2x+8)
=(x+10)/(2x+8)
(a+2/4b) - (a-1/10b) + (a-3/5b)
=[(5a+10) - (2a-2) + (4a-12)]/20b
=(5a+10-2a+2+4a-12) /20b
=7a/20b
2007-06-27 10:45:31 · answer #1 · answered by fofo m 3 · 0⤊ 0⤋
Some of your respondents are making errors in multiplying as it these were equations, they are not.
the correct answer to th first one is -2/b -2. It is still a binomial.
The second one if I understand your grouping results in:
(-6y +49)/(y^2 -49) there is no purpose to factor y^2- 49 since you cannot cancel any of it with any part of the numerator.
In the third problem all you can do is remove the () and collect like terms to get 4.5+ 3/x, still a binomial. Don't make the mistake that you can find a common denominator and then add them up. Unless these all equal something, you cannot multiply thru with a common denominator then cancel.... since these are just expressions, not equations.
The third has the same problem:
It comes out as (-6y + 49)/(y^2-49) It us useless to factor the denominator into (y+7)(y-7) since you can't factor any of the numerator to cancel.
If any of these equaled something, the LCD would work and these would be simpler.
2007-06-27 11:40:43 · answer #2 · answered by April 6 · 0⤊ 0⤋
(b+1/b+2) - (b+3/b+4)
=[(b+1)(b+4)-(b+3)(b+2)]/[(b+2)(b+4)]
=[b^2+5b+4 -b^2 -5b-6]/[(b+2)(b+4)]
= -2/[(b+2)(b+4)]
(y/y^2-49) - (7/y+7)
= y/[(y-7)(y+7)] - 7(y-7)/[(y-7)(y+7)]
=( y-7y+49)/[(y-7)(y+7)]
=(-6y+49)/(y^2-49)
(1/2) + 3/(x+4)
= (x+4 +3*2)/[2(x+4)] = (x+10/(2x+8)
(a+2/4b) - (a-1/10b) + (a-3/5b)
= a+ 10/20b -a - 2/20b +a -12/20b
= a-4/20b = a - 1/5b
2007-06-27 10:52:23 · answer #3 · answered by ironduke8159 7 · 0⤊ 0⤋
1. (b+1/b+2) - (b+3/b+4)
First remove the brackets & apply the sign:
b + 1/b + 2 - b - 3/b - 4
Gather like terms:
b-b + 1/b-3/b - 4
Simplify:
- 2/b - 4
This doesn't look right are you sure of the question?
2007-06-27 10:41:14 · answer #4 · answered by Robert S 7 · 0⤊ 0⤋
1.-2/(b+2)(b+4)
2. -3/28
3.x=-2/3
4. (5ab-1)/5b
2007-06-27 10:32:33 · answer #5 · answered by balas i 2 · 0⤊ 1⤋