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Posted by **ryan** on Thursday, April 25, 2013 at 12:08am.

- algebra1 -
**ryan**, Thursday, April 25, 2013 at 1:18amsolve 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 1:20amEasy x=2 and x=3

- algebra1 -
**ryan**, Thursday, April 25, 2013 at 1:39amsolve 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 dont get these problem porrect.

- algebra1 -
**Reiny**, Thursday, April 25, 2013 at 8:40amrate 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 8:49amPlease 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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