How do you simplify #(k^2-26)/(k-5)-1/(5-k)#?

1 Answer
May 4, 2017

See the solution process below:

Explanation:

To subtract or add and simplify these two fractions they must be over a common denominator. Multiply the fraction on the right by the appropriate form of #1# which is (#(-1)/-1#):

#(k^2 - 26)/(k - 5) - ((-1)/-1 xx 1/(5 - k)) =>#

#(k^2 - 26)/(k - 5) - (-1/(-1(5 - k))) =>#

#(k^2 - 26)/(k - 5) - (-1/((-1 xx 5) - (-1 xx k))) =>#

#(k^2 - 26)/(k - 5) - (-1/(-5 - (-1k))) =>#

#(k^2 - 26)/(k - 5) - (-1/(-5 + 1k))) =>#

#(k^2 - 26)/(k - 5) - (-1/(-5 + k))) =>#

#(k^2 - 26)/(k - 5) - (-1/(k - 5)) =>#

#(k^2 - 26)/(k - 5) + 1/(k - 5)#

We can now add the numerator over the common denominator:

#(k^2 - 26 + 1)/(k - 5)#

#(k^2 - 25)/(k - 5)#

The numerator is a special for of the quadratic:

#a^2 - b^2 = (a + b)(a - b)#

We can factor the remaining fraction as:

#((k + 5)(k - 5))/(k - 5)#

We can now cancel common terms in the numerator and denominator:

#((k + 5)color(red)(cancel(color(black)((k - 5)))))/color(red)(cancel(color(black)(k - 5))) =>#

#k + 5# Where, from the original expression #k != 5#