How do you simplify #7/(4k+8)-k/(k+2)#?

1 Answer
Jun 3, 2017

See a solution process below:

Explanation:

To be able to add or subtract fractions, both fractions must be over a common denominator. In this case #(4k + 8)#. We can put the fraction on the right over this common denominator by multiplying the fraction by the appropriate form of #1#:

#7/(4k + 8) - (4/4 xx k/(k + 2)) =>#

#7/(4k + 8) - (4k)/(4(k + 2)) =>#

#7/(4k + 8) - (4k)/((4 * k) + (4 * 2)) =>#

#7/(4k + 8) - (4k)/(4k + 8)#

We can now subtract the numerators over the common denominator:

#7/(4k + 8) - (4k)/(4k + 8) =>#

#(7 - 4k)/(4k + 8)#