Question 444212
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Hi
 1,2 and 4 Set Up for Cross Multiplying to solve:
{{{x/2 = (x+1)/4}}}   4x = 2(x+1)   2x = 2   x = 1
{{{3/(x-1) = 4/(3x+2)}}}   3(3x+2) = 4(x-1)    5x = -10  x = -2
{{{3/(x-1) = 1/(x^2-1)}}}
   3(x^2-1) = (x-1)
  3x^2 - x -2 = 0
factoring
 (3x+2)(x-1) = 0
 (3x+2)= 0   x = -2/3
  (x-1)= 0   x = 1

3. 3x/(x+1) = 0  x = 0 would work