Question

How do you simplify `(a+2b)/(x+a)-(a-2b)/(x-a)-(4bx-2a^2)/(x^2-a^2)`?

Answer 1

`=0`

Explanation:

`(a+2b)/(x+a)-(a-2b)/(x-a)-(4bx-2a^2)/(x^2-a^2)" "larr` factorise

`(a+2b)/(x+a)-(a-2b)/(x-a)-((4bx-2a^2))/((x+a)(x-a))`

Find a common denominator and equivalent fractions:

`(a+2b)/(x+a)xxcolor(blue)((x-a)/(x-a))-(a-2b)/(x-a)xxcolor(blue)( (x+a)/(x+a))-((4bx-2a^2))/((x+a)(x-a))`

`((a+2b)(x-a) -(a-2b)(x+a) - (4bx-2a^2))/((x+a)(x-a))`

`=((ax-a^2+2bx-2ab)-(ax+a^2-2bx-2ab)-(4bx-2a^2))/((x+a)(x-a))`

`=(color(blue)(ax)color(red)(-a^2)color(forestgreen)(+2bx)color(magenta)(-2ab)color(blue)(-ax)color(red)(-a^2)color(forestgreen)(+2bx)color(magenta)(+2ab)color(forestgreen)(-4bx)color(red)(+2a^2))/((x+a)(x-a))`

`=(color(blue)(0ax)color(red)(+0a^2)color(forestgreen)(+0bx)color(magenta)(+0ab))/((x+a)(x-a))`

`=0`

Answer 2

`0`

Explanation:

`"before we can add/subtract fractions we require them to"`
`"have a "color(blue)"common denominator"`

`x^2-a^2" is a "color(blue)"difference of squares"`

`rArra^2-x^2=(x-a)(x+a)`

`"multiply numerator/denominator of "(a+2b)/(x+a)" by "(x-a)`

`"multiply numerator/denominator of "(a-2b)/(x-a)" by "(x+a)`

`((a+2b)(x-a))/((x-a)(x+a))-((a-2b)(x+a))/((x-a)(x+a))-(4bx-2a^2)/((x-a)(x+a))`

`"now add/subtract the numerators leaving the denominator"`
`"first multiplying out the brackets"`

`(ax+2bx-a^2-2ab-(ax-2bx+a^2-2ab)-(4bx-2a^2))/((x-a)(x+a))`

`=(ax+2bx-a^2-2ab-ax+2bx-a^2+2ab-4bx+2a^2)/((x-a)(x+a))`

`=0`