How do you evaluate #\frac { 3} { x ^ { 2} - 9} + \frac { 11} { 2x + 6} = 1#?

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Answer 1 · 2017-01-04T14:59:53

We first try to get the fractions sorted (factorize)

#->3/((x+3)(x-3))+11/(2(x+3))=1#

The LCM of the denominators is #2(x+3)(x-3)#,
so we multiply everything by that:
(I'll spare you the in-betweens, but a lot cancels out):

#->3xx2+11xx(x-3)=1xx2(x+3)(x-3)#

Work out the brackets:
#->6+11x-33=2x^2-18#

Take everything to one side:
#0=2x^2-11x+9#

Solve this with the ABC-formula:

#x_(1.2)=(-(-11)+-sqrt((-11)^2-4*2*9))/(2*2)#

#x_(1.2)=(11+-sqrt(121-72))/4=11/4+-1/4sqrt49#

#x_(1.2)=11/4+-7/4# So #x_1=18/4=9/2and x_2=4/4=1#