How do you solve #(x+2)/(x+3)-1=1/(3-2x-x^2)# and check for extraneous solutions?

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1 Answer
Douglas K.
May 13, 2017

Please see the explanation.

Explanation:

Given: #(x+2)/(x+3)-1=1/(3-2x-x^2)#

Multiply the right side by 1 in the form of #(-1)/(-1)#:

#(x+2)/(x+3)-1=-1/(x^2+2x-3)#

Factor the right side's denominator:

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

Restrict the domain to prevent division by 0:

#(x+2)/(x+3)-1=-1/((x-1)(x+3)); x!=1, x!=-3#

NOTE: We restrict the values of x to make us notice, if we obtain an extraneous root.

Multiply both sides by #x+3#

#(x+2)-x - 3=-1/(x-1); x!=1, x!=-3#

Combine like terms:

#-1 = -1/(x-1); x!=1, x!=-3#

Multiply both sides by #-(x-1)#:

#x-1 = 1; x!=1, x!=-3#

Add 1 to both sides and drop the restrictions, because the solution will not violate them.

#x = 2#

Check:

#(2+2)/(2+3)-1=1/(3-2(2)-2^2)#

#4/5-1 = 1/-5#

#-1/5=-1/5#

This checks.

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