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

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Answer 1 · 2016-09-04T00:33:31

#a = -19/8#

Put on a common denominator.

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

Eliminate the denominators and solve the resulting quadratic.

#a^2 - 4 - 1(a^2 + 5a + 6) = 3a + 9#

#a^2 - 4 - a^2 - 5a - 6 = 3a + 9#

#-8a = 19#

#a = -19/8#

Hopefully this helps!