How do you simplify # a/(r+a) - a/(r-a)#?

2 Answers
Jim G.
Nov 9, 2017

#-(2a^2)/((r+a)(r-a))#

Explanation:

#"before we can subtract the fractions we require them to"#
#"have a "color(blue)"common denominator"#

#"multiply numerator/denominator of " a/(r+a)" by "(r-a)#

#"multiply numerator/denominator of "a/(r-a)" by "(r+a)#

#rArr(a(r-a))/((r+a)(r-a))-(a(r+a))/((r+a)(r-a))#

#"we now have a common denominator so can subtract"#
#"the numerators leaving the denominator"#

#=(cancel(ar)-a^2cancel(-ar)-a^2)/((r+a)(r-a))#

#=-(2a^2)/((r+a)(r-a))to(r!=+-a)#

smendyka
Nov 9, 2017

See a solution process below:

Explanation:

First, we will put the fractions over common denominators by multiplying each fraction by the appropriate form of #1#:

#((r - a)/(r - a) xx a/(r + a)) - ((r + a)/(r + a) xx a/(r - a))#

#((r - a)a)/((r - a)(r + a)) - ((r + a)a)/((r + a)(r - a))#

#(ar - a^2)/(r^2 + ar - ar - a^2) - (ar + a^2)/(r^2 + ar - ar - a^2)#

#(ar - a^2)/(r^2 - a^2) - (ar + a^2)/(r^2 - a^2)#

We can now subtract the numerators over the common denominators:

#((ar - a^2) - (ar + a^2))/(r^2 - a^2)#

#(ar - a^2 - ar - a^2)/(r^2 - a^2)#

#(ar - ar - a^2 - a^2)/(r^2 - a^2)#

#(0 - 2a^2)/(r^2 - a^2)#

#(-2a^2)/(r^2 - a^2)#

However, we must look at the original expression and qualify the answer with:

Where #r + a != 0# or #r - a != 0#

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