Posted by Unknown on Tuesday, August 21, 2012 at 2:51pm.
You will need to use the distributive property of multiplication.
5/(2y+6) + 1/(y-2) = 3/(y+3)
5/2(y+3) + 1/(y-2) = 3/(y+3)
2(5)/(y+3) + 2/(y-2) = 2(3)/(y+3)
10/(y+3) + 2/(y-2) = 6/(y+3)
2/(y-2) = -4/(y+3)
(y-2)(2)/(y-2) = -4(y-2)/(y+3)
2 = -4(y-2)/(y+3)
2/(-4) = -4(y-2)/-4(y+3)
-(1/2) = (y-2)/(y+3)
-(1/2)(y+3) = (y+3)(y+2)/(y+3)
-(1/2)(y+3) = (y+2)
-(1/2)y - 3/2 = y + 2
-(3/2) - 2 = y + (1/2)y
-(3/2) - 4/2 = (2/2)y + (1/2)y
-(7/2) = (3/2)y
-(7/2) / (3/2) = y
- (7/3) = y
y = - 2 1/3
hmmm. I don't get that.
5/(2y+6) + 1/(y-2) = 3/(y+3)
putting all over 2(y+3)(y-2), and discarding the denominator, we have
5(y-2) + 2(y+3) = 3*2(y-2)
5y - 10 + 2y + 6 = 6y - 12
y = -8
check:
5/-10 + 1/-10 = 6/-10 = 3/-5
----------------------------
3/(3x-2) = 4/(2x+1)
putting all over (3x-2)(2x+1), we have
3(2x+1) = 4(3x-2)
6x + 3 = 12x - 8
6x = 11
x = 11/6
-------------------------------
5/3x - 1/9 = 1/x
put all over 9x:
5*3 - x = 9
x = 6
------------------------------
2/(x^2-x-12) -6/(x+4) = 2/(x-3)
note that x^2-x-12 = (x-4)(x+3)
So, putting all over (x-4)(x+4)(x-3)(x+3), we have
2(x+4)(x-3) - 6(x+4)(x-3)(x+3) = 2(x-4)(x+4)(x+3)
Now, that's kind of nasty. I think you have a typo. If instead we have
2/(x^2+x-12) -6/(x+4) = 2/(x-3)
the LCD = (x+4)(x-3), and we get
2 - 6(x-3) = 2(x+4)
2 - 6x + 18 = 2x + 8
x = 3/2