((x-3))/(3)+((2x+4))/(5) - add fractions

Solution for the given fractions

• (x-3)/3 + (2*x+4)/5 = ?
• The common denominator of the two fractions is: 15
• (x-3)/3 = (5*(x-3))/(3*5) = (5*(x-3))/15
• (2*x+4)/5 = (3*(2*x+4))/(3*5) = (3*(2*x+4))/15
• Fractions adjusted to a common denominator
• (x-3)/3 + (2*x+4)/5 = (5*(x-3))/15 + (3*(2*x+4))/15
• (5*(x-3))/15 + (3*(2*x+4))/15 = (5*(x-3)+3*(2*x+4))/15
• (5*(x-3)+3*(2*x+4))/15 = (5*(x-3)+3*(2*x+4))/15

Solution for the given fractions

$ \frac{(x-3)}{3 }+ \frac{(2*x+4)}{5 }=? $

The common denominator of the two fractions is: 15

$ \frac{(x-3)}{3 }= \frac{(5*(x-3))}{(3*5)} = \frac{(5*(x-3))}{15} $

$ \frac{(2*x+4)}{5 }= \frac{(3*(2*x+4))}{(3*5)} = \frac{(3*(2*x+4))}{15} $

Fractions adjusted to a common denominator

$ \frac{(x-3)}{3 }+ \frac{(2*x+4)}{5 }= \frac{(5*(x-3))}{15 }+ \frac{(3*(2*x+4))}{15} $

$ \frac{(5*(x-3))}{15 }+ \frac{(3*(2*x+4))}{15 }= \frac{(5*(x-3)+3*(2*x+4))}{15} $

$ \frac{(5*(x-3)+3*(2*x+4))}{15 }= \frac{(5*(x-3)+3*(2*x+4))}{15} $

$ $

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