How do you simplify #5/(2k+2)-k/(d+5)#?

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Answer 1 · 2017-07-12T13:24:02

Multiply to achieve a common denominator so the fractions to be added and simplified.

The LCM would be # (2k +2) xx (d+5) # so

# { 5 xx (d+5)/((2k+2)(d+5)) } = k xx (2k+2)/( (2k+2) xx (d+5))}#

This gives

# (5d + 25) / (2kd +10k + 2d +10)+ (2k^2 + 2k)/ (2kd + 10k + 2d + 10)#

adding the fractions gives

# (2k^2 +2k + 5d + 25)/ ( 2kd + 10k +2d + 10)#

This is not real simple

Answer 2 · 2017-07-12T20:05:33

#(5d+25-2k^2-2k)/(2(k+1)(d+5)#

Factorise the denominator.

#5/(2(k+1)) -k/(d+5)#

Find a common denominator.

#(color(white)(....................................))/(2(k+1)(d+5)#

Find equivalent fractions.

#(5(d+5) -2k(k+1))/(2(k+1)(d+5)#

Simplify:

#(5d+25-2k^2-2k)/(2(k+1)(d+5)#