How do you solve #\frac { 1} { c + 3} = \frac { 7} { c ^ { 2} - 9}#?
1 Answer
Aug 31, 2017
Explanation:
#c^2-9 = 7(c+3)# #-># cross-multiply
#c^2-9 = 7c + 21# #-># distribute
#c^2 - 9 - 7c - 21 = 0# #-># bring all terms to one side
#c^2 - 7c - 30 = 0# #-># gather like terms
#(c-10)(c+3) = 0# #-># factor
We now have two possibilities:
#c - 10 = 0# and#c + 3 = 0#
#c = 10# and#c = -3#
We can check our answers by substituting them back into the original problem.
#1/(c+3) = 7/(c^2 - 9)#
#1/(color(blue)10+3) = 7/(color(blue)10^2 - 9)#
#1/13 = 7 / (100 - 9)#
#1/13 = 7/91#
#1/13 = 1/13#
#underlinecolor(white)(xxxxxxxx)#
#1/(color(blue)(-3) + 3) = 7/(color(blue)((-3))^2-9#
#1/0 = 7/0# This is undefined, so
#c=-3# is not a solution.
Thus,
