Posted by ryan on Thursday, April 25, 2013 at 12:08am.
the school bought a new copy machine which makes 20 copies per minute.the old machine makes 12 copies per minute.francine needs 1000 copies,so she uses both machines.how long will it take to have the copies

algebra1  ryan, Thursday, April 25, 2013 at 1:18am
solve the following system of equations by first rearranging them so that they are not in fractional form:
3/(x+2) 2/(x+3)=7y/(3x^2+15x+18)
4(y+3) +1/(y+1)= 11x/(Y^2+4y+3)

algebra1  Liv4life, Thursday, April 25, 2013 at 1:20am
Easy x=2 and x=3

algebra1  ryan, Thursday, April 25, 2013 at 1:39am
solve the following system of equations by first rearranging them so that they are not in fractional form:
3/(x+2) 2/(x+3)=7y/(3x^2+15x+18)
4(y+3) +1/(y+1)= 11x/(Y^2+4y+3) oh please help oh please my math teach will funk me if i don't get these problem porrect.

algebra1  Reiny, Thursday, April 25, 2013 at 8:40am
rate of 1st machine = 20 copies/min
rate of 2nd machine = 12 copies/min
combined rate = 32 copies/min
so to make 1000 copies takes
1000/32 or 31.25 minutes or
31 minutes and 15 seconds

Note to Ryan  algebra1  Reiny, Thursday, April 25, 2013 at 8:49am
Please do not post a new question as an answer to an existing problem, it could easily be considered to be an answer and could thus be ignored.
3/(x+2) 2/(x+3)=7y/(3x^2+15x+18)
3/(x+2)  2/(x+3) = 7y/(3(x+2)(x+3))
multiplyeach term by 3(x+2)(x+3) , the LCD
9(x+3)  6(x+2) = 7y
9x + 27  6x  12 = 7y
3x  7y = 15 > #1
Repeat the same thing for the 2nd equation, it works out just as simple
Note that y^2 + 4y + 3 = (y+1)(y+3)
Now you have 2 simple linear equations with x and y, and judging by the complexity of the two starting equations, you obviously know how to do that.
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