Assuming this is \(\displaystyle \frac{yx+ 3}{4x- 2}\), since both numerator and denominator are of first degree, "long division" will give a number plus a fraction with denominator 4x-2. In order that this "not depend on x", the numerator of that fraction (the "remainder" when yx+ 3 is divided by 4x- 2) must be 0.
You could also do this by noting that 4x- 2= 4(x- 1/2) so that \(\displaystyle \frac{yx+ 3}{4x- 2}= \frac{1}{4}\frac{yx+ 3}{x- 1/2}\) and the remainder when polynomial, p(x) is divided by x- 1/2 is p(1/2).