(2x-7)/(3x)+(4x-9)/(6) - adding of fractions

Solution for the given fractions

• (2*x-7)/(3*x) + (4*x-9)/6 = ?
• The common denominator of the two fractions is: 18*x
• (2*x-7)/(3*x) = (6*(2*x-7))/(3*6*x) = (6*(2*x-7))/(18*x)
• (4*x-9)/6 = (3*x*(4*x-9))/(3*6*x) = (3*x*(4*x-9))/(18*x)
• Fractions adjusted to a common denominator
• (2*x-7)/(3*x) + (4*x-9)/6 = (6*(2*x-7))/(18*x) + (3*x*(4*x-9))/(18*x)
• (6*(2*x-7))/(18*x) + (3*x*(4*x-9))/(18*x) = (6*(2*x-7)+3*x*(4*x-9))/(18*x)
• (6*(2*x-7)+3*x*(4*x-9))/(18*x) = (6*(2*x-7)+3*x*(4*x-9))/(18*x)

Solution for the given fractions

$ \frac{(2*x-7)}{(3*x)} + \frac{(4*x-9)}{6 }=? $

The common denominator of the two fractions is: 18*x

$ \frac{(2*x-7)}{(3*x)} = \frac{(6*(2*x-7))}{(3*6*x)} = \frac{(6*(2*x-7))}{(18*x)} $

$ \frac{(4*x-9)}{6 }= \frac{(3*x*(4*x-9))}{(3*6*x)} = \frac{(3*x*(4*x-9))}{(18*x)} $

Fractions adjusted to a common denominator

$ \frac{(2*x-7)}{(3*x)} + \frac{(4*x-9)}{6 }= \frac{(6*(2*x-7))}{(18*x)} + \frac{(3*x*(4*x-9))}{(18*x)} $

$ \frac{(6*(2*x-7))}{(18*x)} + \frac{(3*x*(4*x-9))}{(18*x)} = \frac{(6*(2*x-7)+3*x*(4*x-9))}{(18*x)} $

$ \frac{(6*(2*x-7)+3*x*(4*x-9))}{(18*x)} = \frac{(6*(2*x-7)+3*x*(4*x-9))}{(18*x)} $

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