How do you combine #(x-2)/(x-6)-(x+2)/(6-x)#?
1 Answer
Explanation:
The trick here is to realize that you can write
#6 -x = -1 * (-6 + x) = -1 * (x-6)#
This means that the expression can be rewritten as
#(x-2)/(x-6) - (x+2)/(-1 * (x-6)) = (x-2)/(x-6) - (x+2)/(-(x-6))#
This will be equivalent to
#(x-2)/(x-6) + (x+2)/(x-6)#
You now have two fractions that have the same denominator, which means that you can go ahead and combine their numerators
#(x-2)/(x-6) + (x+2)/(x-6) = (x - color(red)(cancel(color(black)(2))) + x + color(red)(cancel(color(black)(2))))/(x-6) = color(green)(|bar(ul(color(white)(a/a)color(black)((2x)/(x-6))color(white)(a/a)|)))#
ALTERNATIVELY
You can also combine these two fractions by finding their common denominator, which in this case would be
Multiply the first fraction by
#(x-2)/(x-6) - (x+2)/(6-x) = (x-2)/(x-6) * (6-x)/(6-x) - (x+2)/(6-x) * (6-x)/(6-x)#
#=((x-2)(6-x))/((x-6)(6-x)) - ((x+2)(6-x))/((x-6)(6-x))#
#= ((x-2)(6-x) - (x+2)(x-6))/((x-6)(6-x))#
Focus on the numerator first
#6x - x^2 - color(red)(cancel(color(black)(12))) + color(red)(cancel(color(black)(2x))) - x^2 + 6x - color(red)(cancel(color(black)(2x))) + color(red)(cancel(color(black)(12))) = -2x^2 + 12x#
This can be rewritten as
#-2x^2 + 12x = 12x - 2x^2 = 2x * (6 - x)#
The expression will once again be equal to
#(2x * color(red)(cancel(color(black)((6-x)))))/((x-6)color(red)(cancel(color(black)((6-x))))) = color(green)(|bar(ul(color(white)(a/a)color(black)((2x)/(x-6))color(white)(a/a)|)))#
Keep in mind that you need to have