(3x+6)/(10)+(2x-4)/(4) - adding of fractions

Solution for the given fractions

• (3*x+6)/10 + (2*x-4)/4 = ?
• The common denominator of the two fractions is: 20
• (3*x+6)/10 = (2*(3*x+6))/(2*10) = (2*(3*x+6))/20
• (2*x-4)/4 = (5*(2*x-4))/(4*5) = (5*(2*x-4))/20
• Fractions adjusted to a common denominator
• (3*x+6)/10 + (2*x-4)/4 = (2*(3*x+6))/20 + (5*(2*x-4))/20
• (2*(3*x+6))/20 + (5*(2*x-4))/20 = (2*(3*x+6)+5*(2*x-4))/20
• (2*(3*x+6)+5*(2*x-4))/20 = (2*(3*x+6)+5*(2*x-4))/20

Solution for the given fractions

$ \frac{(3*x+6)}{10 }+ \frac{(2*x-4)}{4 }=? $

The common denominator of the two fractions is: 20

$ \frac{(3*x+6)}{10 }= \frac{(2*(3*x+6))}{(2*10)} = \frac{(2*(3*x+6))}{20} $

$ \frac{(2*x-4)}{4 }= \frac{(5*(2*x-4))}{(4*5)} = \frac{(5*(2*x-4))}{20} $

Fractions adjusted to a common denominator

$ \frac{(3*x+6)}{10 }+ \frac{(2*x-4)}{4 }= \frac{(2*(3*x+6))}{20 }+ \frac{(5*(2*x-4))}{20} $

$ \frac{(2*(3*x+6))}{20 }+ \frac{(5*(2*x-4))}{20 }= \frac{(2*(3*x+6)+5*(2*x-4))}{20} $

$ \frac{(2*(3*x+6)+5*(2*x-4))}{20 }= \frac{(2*(3*x+6)+5*(2*x-4))}{20} $

$ $

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