How do you simplify 2x+3/(x^2-9) + x/(x-3)2x+3x2−9+xx−3?
1 Answer
Jan 11, 2018
Explanation:
"before adding the fractions we require them to have"before adding the fractions we require them to have
"a "color(blue)"common denominator"a common denominator
"factorise the denominator "x^2-9factorise the denominator x2−9
x^2-9=(x-3)(x+3)larrcolor(blue)"difference of squares"x2−9=(x−3)(x+3)←difference of squares
"multiply numerator/denominator of"x/(x-3)" by "(x+3)multiply numerator/denominator ofxx−3 by (x+3)
=3/((x-3)(x+3))+(x(x+3))/((x-3)(x+3))=3(x−3)(x+3)+x(x+3)(x−3)(x+3)
"add the numerators leaving the denominator"add the numerators leaving the denominator
=(3+x^2+3x)/((x-3)(x+3))=(x^2+3x+3)/((x-3)(x+3))=3+x2+3x(x−3)(x+3)=x2+3x+3(x−3)(x+3)
