How do you simplify and find the restrictions for #(3x-2)/(x+3)+7/(x^2-x-12)#?

Question

Savannah Anderson · Accepted Answer

#=((x-3)(3x-5))/((x-4)(x+3)#
Restrictions are
#x!=4#
#x!=-3# #(3x-2)/(x+3)+7/(x^2-x-12)# #=(3x-2)/(x+3)+7/(x^2-4x+3x-12)# #=(3x-2)/(x+3)+7/(x(x-4)+3(x-4)# #=(3x-2)/(x+3)+7/((x-4)(x+3))# #=((3x-2)(x-4)+7)/((x-4)(x+3))# #=(3x^2-12x-2x+8+7)/((x-4)(x+3))# #=(3x^2-14x+15)/((x-4)(x+3))# #=(3x^2-9x-5x+15)/((x-4)(x+3))# #=(3x(x-3)-5(x-3))/((x-4)(x+3))# #=((x-3)(3x-5))/((x-4)(x+3)# Restrictions are #x-4!=0# or #x!=4# #x+3!=0# or #x!=-3#

Ethan Adkins · Answer

undefined To simplify the expression (3x-2)/(x+3)+7/(x^2-x-12), we first need to find the common denominator. The common denominator is (x+3)(x-4).

Next, we can rewrite the expression with the common denominator: [(3x-2)(x-4) + 7(x+3)] / [(x+3)(x-4)].

Expanding and combining like terms in the numerator gives: (3x^2 - 14x + 8 + 7x + 21) / [(x+3)(x-4)].

Simplifying further, we have: (3x^2 - 7x + 29) / [(x+3)(x-4)].

To find the restrictions, we need to identify any values of x that would make the denominator equal to zero. In this case, the restrictions are x = -3 and x = 4, as they would result in division by zero.