How do you solve #\frac { 2} { a ^ { 2} + 3a - 18} + \frac { 1} { a - 3} = \frac { 1} { 2a - 6}#?

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Answer 1 · 2017-12-04T00:13:09

#a=-10#

#2/(a^2+3a-18)+1/(a-3)=1/(2a-6)#

#2/[(a+6)*(a-3)]+1/(a-3)=1/(2a-6)#

#[1*(a+6)+2]/[(a+6)*(a-3)]=1/(2a-6)#

#(a+8)/[(a+6)*(a-3)]=1/[2(a-3)]#

#(a+8)/(a+6)=1/2#

#2*(a+8)=1*(a+6)#

#2a+16=a+6#

#a=-10#