How do you simplify 5/(2k+2)-k/(d+5)52k+2−kd+5?

2 Answers
David Drayer
Jul 12, 2017

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

Explanation:

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

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

This gives

(5d + 25) / (2kd +10k + 2d +10)+ (2k^2 + 2k)/ (2kd + 10k + 2d + 10)5d+252kd+10k+2d+10+2k2+2k2kd+10k+2d+10

adding the fractions gives

(2k^2 +2k + 5d + 25)/ ( 2kd + 10k +2d + 10)2k2+2k+5d+252kd+10k+2d+10

This is not real simple

EZ as pi
Jul 12, 2017

(5d+25-2k^2-2k)/(2(k+1)(d+5)5d+25−2k2−2k2(k+1)(d+5)

Explanation:

Factorise the denominator.

5/(2(k+1)) -k/(d+5)52(k+1)−kd+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)

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