How do you solve #(-1)/(x-3)=(x-4)/(x^2-27)#?

2 Answers
Jul 30, 2016

Answer:

#x=-3/2# or #x=5#

Explanation:

In #(-1)/(x-3)=(x-4)/(x^2-27)#, let us multiply each side by #(x-3)(x^2-27)# to simplify we get

#(-1)/(x-3)xx(x-3)(x^2-27)=(x-4)/(x^2-27)xx(x-3)(x^2-27)# or

#(-1)/cancel((x-3))xxcancel(x-3)(x^2-27)=(x-4)/cancel(x^2-27)xx(x-3)cancel((x^2-27))# or

#-1xx(x^2-27)=(x-4)(x-3)# or

#-x^2+27=x(x-3)-4(x-3)# or

#-x^2+27=x^2-3x-4x+12# or

#-x^2+27-x^2+7x-12=0# or

#-2x^2+7x+15=0#

#2x^2-7x-15=0#

#2x^2-10x+3x-15=0# or

#2x(x-5)+3(x-5)=0# or

#(2x+3)(x-5)=0# and hence

ether #2x+3=0# i.e. #x=-3/2#

or #x-5=0# i.e. #x=5#

Jul 30, 2016

Answer:

#x=5#
#x=-3/2#

Explanation:

#-1/(x-3)=(x-4)/(x^2-27#
or
#-x^2+27=(x-3)(x-4)#
or
#-x^2+27=x^2-3x-4x+12#
or
#-x^2+27=x^2-7x+12#
or
#2x^2-7x-15=0#
or
#2x^2-10x+3x-15=0#
or
#2x(x-5)+3(x-5)=0#
or
#(x-5)(2x+3)=0#
or
#x-5=0#
or
#x=5#=======Ans #1#
or
#2x+3=0#
or
#2x=-3#
or
#x=-3/2#=======Ans #2#