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

1 Answer
Jul 3, 2017

Answer:

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#