(6d-5)/(8d)-(3d-4)/(6d) - subtract fractions

Solution for the given fractions

• (6*d-5)/(8*d) - (3*d-4)/(6*d) = ?
• The common denominator of the two fractions is: 48*d^2
• (6*d-5)/(8*d) = (6*d*(6*d-5))/(6*8*d*d) = (6*d*(6*d-5))/(48*d^2)
• (3*d-4)/(6*d) = (8*d*(3*d-4))/(6*8*d*d) = (8*d*(3*d-4))/(48*d^2)
• Fractions adjusted to a common denominator
• (6*d-5)/(8*d) - (3*d-4)/(6*d) = (6*d*(6*d-5))/(48*d^2) - (8*d*(3*d-4))/(48*d^2)
• (6*d*(6*d-5))/(48*d^2) - (8*d*(3*d-4))/(48*d^2) = (6*d*(6*d-5)-(8*d*(3*d-4)))/(48*d^2)
• (6*d*(6*d-5)-(8*d*(3*d-4)))/(48*d^2) = (6*d*(6*d-5)-8*d*(3*d-4))/(48*d^2)

Solution for the given fractions

$ \frac{(6*d-5)}{(8*d)} - \frac{(3*d-4)}{(6*d)} =? $

The common denominator of the two fractions is: 48*d^2

$ \frac{(6*d-5)}{(8*d)} = \frac{(6*d*(6*d-5))}{(6*8*d*d)} = \frac{(6*d*(6*d-5))}{(48*d^2)} $

$ \frac{(3*d-4)}{(6*d)} = \frac{(8*d*(3*d-4))}{(6*8*d*d)} = \frac{(8*d*(3*d-4))}{(48*d^2)} $

Fractions adjusted to a common denominator

$ \frac{(6*d-5)}{(8*d)} - \frac{(3*d-4)}{(6*d)} = \frac{(6*d*(6*d-5))}{(48*d^2)} - \frac{(8*d*(3*d-4))}{(48*d^2)} $

$ \frac{(6*d*(6*d-5))}{(48*d^2)} - \frac{(8*d*(3*d-4))}{(48*d^2)} = \frac{(6*d*(6*d-5)-(8*d*(3*d-4)))}{(48*d^2)} $

$ \frac{(6*d*(6*d-5)-(8*d*(3*d-4)))}{(48*d^2)} = \frac{(6*d*(6*d-5)-8*d*(3*d-4))}{(48*d^2)} $

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