[tex]x = \dfrac{1}{4}x+\dfrac{3}{4}\ \ \ \ |-\dfrac{1}{4}x\\\\\dfrac{3}{4}x=\dfrac{3}{4}\ \ \ \text{ONE SOLUTION}\\\\\dfrac{1}{2}(1+4x)=2x-3\\\\\dfrac{1}{2}+2x=2x-3\ \ \ \ |-2x\\\\\dfrac{1}{2}=-3\ \ \ \ \text{NO SOLUTION}\\\\\dfrac{1}{3}x-5=\dfrac{2}{3}x-5\ \ \ \ |-\dfrac{2}{3}x\\\\-\dfrac{1}{3}x-5=-5\ \ \ \ \ \text{ONE SOLUTION}\\\\3-4x=-6\left(\dfrac{2}{3}x-\dfrac{1}{2}\right)\\\\3-4x=-4x+3\ \ \ \ |+4x\\\\3=3\ \ \ \ \ \text{INFINITELY MANY SOLUTIONS}[/tex]
