[tex]p-\dfrac{1}{4}=\dfrac{3}{8}+\dfrac{1}{2}p\ \ \ \ |\cdot8\\\\8p-\not8^2\cdot\dfrac{1}{\not4_1}=\not8^1\cdot\dfrac{3}{\not8_1}+\not8^4\cdot\dfrac{1}{\not2_1}p\\\\8p-2=3+4p\ \ \ \ |+2\\\\8p=5+4p\ \ \ \ |-4p\\\\4p=5\ \ \ \ |:4\\\\\boxed{p=\dfrac{5}{4}}[/tex]
