algebra1
posted by ryan .
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
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
Easy x=2 and x=3

algebra1 
ryan
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
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
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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