(4)/(z+1)+(3)/(3z+3) - adding of fractions

Solution for the given fractions

• 4/(z+1) + 3/(3*z+3) = ?
• The common denominator of the two fractions is: (z+1)*(3*z+3)
• 4/(z+1) = (4*(3*z+3))/((z+1)*(3*z+3)) = (4*(3*z+3))/((z+1)*(3*z+3))
• 3/(3*z+3) = (3*(z+1))/((3*z+3)*(z+1)) = (3*(z+1))/((z+1)*(3*z+3))
• Fractions adjusted to a common denominator
• 4/(z+1) + 3/(3*z+3) = (4*(3*z+3))/((z+1)*(3*z+3)) + (3*(z+1))/((z+1)*(3*z+3))
• (4*(3*z+3))/((z+1)*(3*z+3)) + (3*(z+1))/((z+1)*(3*z+3)) = (4*(3*z+3)+3*(z+1))/((z+1)*(3*z+3))
• (4*(3*z+3)+3*(z+1))/((z+1)*(3*z+3)) = (4*(3*z+3)+3*(z+1))/((z+1)*(3*z+3))

Solution for the given fractions

$ \frac{4}{(z+1)} +\frac{ 3}{(3*z+3)} =? $

The common denominator of the two fractions is: (z+1)*(3*z+3)

$ \frac{4}{(z+1)} = \frac{(4*(3*z+3))}{((z+1)*(3*z+3))} = \frac{(4*(3*z+3))}{((z+1)*(3*z+3))} $

$ \frac{3}{(3*z+3)} = \frac{(3*(z+1))}{((3*z+3)*(z+1))} = \frac{(3*(z+1))}{((z+1)*(3*z+3))} $

Fractions adjusted to a common denominator

$ \frac{4}{(z+1)} +\frac{ 3}{(3*z+3)} = \frac{(4*(3*z+3))}{((z+1)*(3*z+3))} + \frac{(3*(z+1))}{((z+1)*(3*z+3))} $

$ \frac{(4*(3*z+3))}{((z+1)*(3*z+3))} + \frac{(3*(z+1))}{((z+1)*(3*z+3))} = \frac{(4*(3*z+3)+3*(z+1))}{((z+1)*(3*z+3))} $

$ \frac{(4*(3*z+3)+3*(z+1))}{((z+1)*(3*z+3))} = \frac{(4*(3*z+3)+3*(z+1))}{((z+1)*(3*z+3))} $

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