(1y)/(2)-(1)/(4y) - subtract fractions

Solution for the given fractions

• (1*y)/2 - 1/(4*y) = ?
• The common denominator of the two fractions is: 8*y
• (1*y)/2 = (1*4*y*y)/(2*4*y) = (4*y^2)/(8*y)
• 1/(4*y) = (1*2)/(2*4*y) = 2/(8*y)
• Fractions adjusted to a common denominator
• (1*y)/2 - 1/(4*y) = (4*y^2)/(8*y) - 2/(8*y)
• (4*y^2)/(8*y) - 2/(8*y) = (4*y^2-2)/(8*y)
• (4*y^2-2)/(8*y) = (4*y^2-2)/(8*y)

Solution for the given fractions

$ \frac{(1*y)}{2 }-\frac{ 1}{(4*y)} =? $

The common denominator of the two fractions is: 8*y

$ \frac{(1*y)}{2 }= \frac{(1*4*y*y)}{(2*4*y)} = \frac{(4*y^2)}{(8*y)} $

$ \frac{1}{(4*y)} = \frac{(1*2)}{(2*4*y)} =\frac{ 2}{(8*y)} $

Fractions adjusted to a common denominator

$ \frac{(1*y)}{2 }-\frac{ 1}{(4*y)} = \frac{(4*y^2)}{(8*y)} -\frac{ 2}{(8*y)} $

$ \frac{(4*y^2)}{(8*y)} -\frac{ 2}{(8*y)} = \frac{(4*y^2-2)}{(8*y)} $

$ \frac{(4*y^2-2)}{(8*y)} = \frac{(4*y^2-2)}{(8*y)} $

$ $

see mathematical notation