(x+1)/(5)+(48+6x)/(32x) - adding of fractions

Solution for the given fractions

• (x+1)/5 + (6*x+48)/(32*x) = ?
• The common denominator of the two fractions is: 160*x
• (x+1)/5 = (32*x*(x+1))/(5*32*x) = (32*x*(x+1))/(160*x)
• (6*x+48)/(32*x) = (5*(6*x+48))/(5*32*x) = (5*(6*x+48))/(160*x)
• Fractions adjusted to a common denominator
• (x+1)/5 + (6*x+48)/(32*x) = (32*x*(x+1))/(160*x) + (5*(6*x+48))/(160*x)
• (32*x*(x+1))/(160*x) + (5*(6*x+48))/(160*x) = (32*x*(x+1)+5*(6*x+48))/(160*x)
• (32*x*(x+1)+5*(6*x+48))/(160*x) = (32*x*(x+1)+5*(6*x+48))/(160*x)

Solution for the given fractions

$ \frac{(x+1)}{5 }+ \frac{(6*x+48)}{(32*x)} =? $

The common denominator of the two fractions is: 160*x

$ \frac{(x+1)}{5 }= \frac{(32*x*(x+1))}{(5*32*x)} = \frac{(32*x*(x+1))}{(160*x)} $

$ \frac{(6*x+48)}{(32*x)} = \frac{(5*(6*x+48))}{(5*32*x)} = \frac{(5*(6*x+48))}{(160*x)} $

Fractions adjusted to a common denominator

$ \frac{(x+1)}{5 }+ \frac{(6*x+48)}{(32*x)} = \frac{(32*x*(x+1))}{(160*x)} + \frac{(5*(6*x+48))}{(160*x)} $

$ \frac{(32*x*(x+1))}{(160*x)} + \frac{(5*(6*x+48))}{(160*x)} = \frac{(32*x*(x+1)+5*(6*x+48))}{(160*x)} $

$ \frac{(32*x*(x+1)+5*(6*x+48))}{(160*x)} = \frac{(32*x*(x+1)+5*(6*x+48))}{(160*x)} $

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