question 1:

[ (2y-1)/2 ] / [ (1-4y^2)/4 ]

well you can see that,

1-4y^2

= 1^2 - (2y)^2

= (1 - 2y)(1 + 2y) =====> (x - a)(x + a) = x^2 - a^2

substituting back to the main question

[ (2y-1)/2 ] / [ (1-2y)(1+2y)/4 ]

= [ (2y-1)/2 ] / [ (-1)(2y-1)(1+2y)/4 ]

= [ 4.(2y-1)/2 ] / [ (-1)(2y-1)(1+2y) ] =====> cancel out the (2y-1) term on the numerator and denominator

= 2 / [ (-1)(1+2y) ]

= -2 / (1+2y)

Question 2:

4/(x-1) = 5/(2x-2) + 3x/4

4/(x-1) = 5/2(x-1) + 3x/4 ======> remove the 2 from (2x-2)

4/(x-1) = (5/2)/(x-1) + 3x/4

(4-5/2) / (x-1) = 3x/4

(3/2) / (x-1) = 3x/4

4(3/2) = (x-1)(3x)

6 = 3x^2 - 3x

3x^2 - 3x - 6 = 0 ======> rearranging

3( x^2 - x - 2) = 0

x^2 - x - 2 = 0

(x-2)(x+1) = 0 ========> factorize the 2nd order ODE

x-2 = 0, x+1 = 0

x=2, x= -1