(4b-10y)/(8b)+(6b-3y)/(8b) - add fractions

Solution for the given fractions

• (4*b-(10*y))/(8*b) + (6*b-(3*y))/(8*b) = ?
• The common denominator of the two fractions is: 64*b^2
• (4*b-(10*y))/(8*b) = (8*b*(4*b-(10*y)))/(8*8*b*b) = (8*b*(4*b-10*y))/(64*b^2)
• (6*b-(3*y))/(8*b) = (8*b*(6*b-(3*y)))/(8*8*b*b) = (8*b*(6*b-3*y))/(64*b^2)
• Fractions adjusted to a common denominator
• (4*b-(10*y))/(8*b) + (6*b-(3*y))/(8*b) = (8*b*(4*b-10*y))/(64*b^2) + (8*b*(6*b-3*y))/(64*b^2)
• (8*b*(4*b-10*y))/(64*b^2) + (8*b*(6*b-3*y))/(64*b^2) = (8*b*(4*b-10*y)+8*b*(6*b-3*y))/(64*b^2)
• (8*b*(4*b-10*y)+8*b*(6*b-3*y))/(64*b^2) = (8*b*(4*b-10*y)+8*b*(6*b-3*y))/(64*b^2)

Solution for the given fractions

$ \frac{(4*b-(10*y))}{(8*b)} + \frac{(6*b-(3*y))}{(8*b)} =? $

The common denominator of the two fractions is: 64*b^2

$ \frac{(4*b-(10*y))}{(8*b)} = \frac{(8*b*(4*b-(10*y)))}{(8*8*b*b)} = \frac{(8*b*(4*b-10*y))}{(64*b^2)} $

$ \frac{(6*b-(3*y))}{(8*b)} = \frac{(8*b*(6*b-(3*y)))}{(8*8*b*b)} = \frac{(8*b*(6*b-3*y))}{(64*b^2)} $

Fractions adjusted to a common denominator

$ \frac{(4*b-(10*y))}{(8*b)} + \frac{(6*b-(3*y))}{(8*b)} = \frac{(8*b*(4*b-10*y))}{(64*b^2)} + \frac{(8*b*(6*b-3*y))}{(64*b^2)} $

$ \frac{(8*b*(4*b-10*y))}{(64*b^2)} + \frac{(8*b*(6*b-3*y))}{(64*b^2)} = \frac{(8*b*(4*b-10*y)+8*b*(6*b-3*y))}{(64*b^2)} $

$ \frac{(8*b*(4*b-10*y)+8*b*(6*b-3*y))}{(64*b^2)} = \frac{(8*b*(4*b-10*y)+8*b*(6*b-3*y))}{(64*b^2)} $

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