Question

How do you simplify and restricted value of `((a - 4) (a + 3)) / (a^2 - a) div( 2 (a - 4)) /( (a + 3) (a - 1))`?

Answer 1

See a solution process below:

Explanation:

First rewrite the denominator of the fraction on the left as:

`((a - 4)(a + 3))/(a(a - 1)) -: (2(a - 4))/((a +3)(a - 1))`

Next, rewrite the expression as:

`(((a - 4)(a + 3))/(a(a - 1)))/((2(a - 4))/((a +3)(a - 1)))`

Now, use this rule for dividing fractions:

`(color(red)(a)/color(blue)(b))/(color(green)(c)/color(purple)(d)) = (color(red)(a) xx color(purple)(d))/(color(blue)(b) xx color(green)(c))`

#(color(red)((a - 4)(a + 3))/color(blue)(a(a - 1)))/(color(green)(2(a - 4))/color(purple)((a + 3)(a - 1))) = (color(red)((a - 4)(a + 3)) xx color(purple)((a + 3)(a - 1)))/(color(blue)(a(a - 1)) xx color(green)(2(a - 4)))#

Next, cancel common terms in the numerator and denominator:

`(color(red)(cancel((a - 4))(a + 3)) xx color(purple)((a + 3)cancel((a - 1))))/(color(blue)(acancel((a - 1))) xx color(green)(2cancel((a - 4)))) =`

`(a + 3)^2/(2a) =`

`(a^2 + 6a + 9)/(2a)`

Excluded values for:

`a(a - 1)` are: `a != 0` and `a != 1`

`(a + 3)(a - 1)` are: `a != -3` and `a != 1`

`2a` are `a != 0`

Therefore the excluded values are: `0, 1, -3`