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

Question

Kennedy Adams · Accepted Answer

#(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)#

Nathan Axel · Answer

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

Julia Adams · Answer

undefined To simplify the expression 5/(2k+2) - k/(d+5), we need to find a common denominator for the two fractions. The common denominator is (2k+2)(d+5).

Multiplying the first fraction by (d+5)/(d+5) and the second fraction by (2k+2)/(2k+2), we get:

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

Expanding the numerators, we have:

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

Combining the fractions, we get:

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

Simplifying the numerator, we have:

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

Therefore, the simplified expression is (5d - 2k^2 - 2k + 25)/((2k+2)(d+5)).