How do you solve #4/(x-5)+2/(x+5)=46/(x^2-25)#?

2 Answers
Ratnaker Mehta
Aug 21, 2016

#+-5# are extraneous solns., as they make the eqn. undefined, whereas, #x=6# is the soln.

Explanation:

We notice that #2# is common throughout the eqn. Hence, we

divide the eqn. by #2#, and, rewrite it as,

#2/(x-5)+1/(x+5)=23/(x^2-25)#.

#rArr {2(x+5)+(x-5)}/{(x-5)(x+5)}=23/(x^2-25)#.

#rArr (3x+5)/(x^2-25)=23/(x^2-25)#.

#rArr (3x+5)(x^2-25)=23(x^2-25)#.

#rArr (3x+5)(x^2-25)-23(x^2-25)=0#.

#rArr (x^2-25)(3x+5-23)=0#.

#rArr (x-5)(x+5)(3x-18)=0#.

#rArr 3(x-5)(x+5)(x-6)=0#.

#rArr x=+-5, x=6#.

Of these, #+-5# are extraneous solns., as they make the eqn.

undefined, whereas, #x=6# is the soln., as it satisfy the eqn.

Deepak G.
Aug 21, 2016

#x=6#

Explanation:

#4/(x-5)+2/(x+5)=46/(x^2-25)#

or

#(4(x+5)+2(x-5))/((x-5)(x+5))=46/(x^2-25)#

or

#(4x+20+2x-10)/((x-5)(x+5))=46/(x^2-25)#

or

#(4x+20+2x-10)/(x^2-25)=46/(x^2-25)#

or

#(6x+10)/(x^2-25)=46/(x^2-25)#

or

#(6x+10)/cancel(x^2-25)=46/cancel(x^2-25)#

or

#6x+10=46#

or

#6x=46-10#

or

#6x=36#

or

#x=36/6#

or

#x=6#

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