Math
posted by Unknown .
Help Solving These?
There Are more like these ,
but i need examples to help me with the others.
1.5/2y+6 + 1/y2=3/y+3
2. 3/3x2 = 4/2x+1
3.5/3x  1/9 = 1/x
4.2/x^2x12 6/x+4 = 2/x3

You will need to use the distributive property of multiplication.
5/(2y+6) + 1/(y2) = 3/(y+3)
5/2(y+3) + 1/(y2) = 3/(y+3)
2(5)/(y+3) + 2/(y2) = 2(3)/(y+3)
10/(y+3) + 2/(y2) = 6/(y+3)
2/(y2) = 4/(y+3)
(y2)(2)/(y2) = 4(y2)/(y+3)
2 = 4(y2)/(y+3)
2/(4) = 4(y2)/4(y+3)
(1/2) = (y2)/(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/(y2) = 3/(y+3)
putting all over 2(y+3)(y2), and discarding the denominator, we have
5(y2) + 2(y+3) = 3*2(y2)
5y  10 + 2y + 6 = 6y  12
y = 8
check:
5/10 + 1/10 = 6/10 = 3/5

3/(3x2) = 4/(2x+1)
putting all over (3x2)(2x+1), we have
3(2x+1) = 4(3x2)
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^2x12) 6/(x+4) = 2/(x3)
note that x^2x12 = (x4)(x+3)
So, putting all over (x4)(x+4)(x3)(x+3), we have
2(x+4)(x3)  6(x+4)(x3)(x+3) = 2(x4)(x+4)(x+3)
Now, that's kind of nasty. I think you have a typo. If instead we have
2/(x^2+x12) 6/(x+4) = 2/(x3)
the LCD = (x+4)(x3), and we get
2  6(x3) = 2(x+4)
2  6x + 18 = 2x + 8
x = 3/2
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