How do you solve #(2a-5)/(a-9)-(a-3)/(3a+2)=5/(3a^2-25a-18)#?

1 Answer
Oct 20, 2016

#a=-3# OR #a=14/5#

Explanation:

#(2a-5)/(a-9)-(a-3)/(3a+2)=5/(3a^2-25a-18)#

#rArr((2a-5)color(blue)((3a+2))-(a-3)color(blue)((a-9)))/color(blue)((a-9)(3a+2))=5/(3a^2-25a-18)#

#rArr((6a^2+4a-15a-10)-(a^2-9a-3a+27))/color(blue)((a-9)(3a+2))=5/(3a^2-25a-18)#

#rArr(6a^2+4a-15a-10-a^2+9a+3a-27)/(3a^2+2a-27a-18)=5/(3a^2-25a-18)#

#rArr(6a^2-a^2+4a-15a+9a+3a-10-27)/(3a^2-25a-18)=5/(3a^2-25a-18)#

#rArr(5a^2+a-37)/(3a^2-25a-18)=5/(3a^2-25a-18)#

#rArr5a^2+a-37=5#
#rArr5a^2+a-37-5=0#
#color(brown)(rArr5a^2+a-42=0)#

Let us compute the quadratic roots based on the property:
#color(red)(delta=b^2-4ac)#
Roots are:
#color(red)(a_1=(-b-sqrtdelta)/(2*a)#
#color(red)(a_2=(-b+sqrtdelta)/(2*a))#

#delta=b^2-4ac=1^2-4(5)(-42)=1+840=841#

roots are:
#a_1=(-1-sqrt841)/(2*5)=(-1-29)/10=(-30/10)=-3#
#a_2=(-1+sqrt841)/(2*5)=(-1+29)/10=(28/10)=14/5#

#color(brown)(rArr5a^2+a-42=0)#
#rArr(a-(color(red)-3))(a-color(red)(14/5))=0#
#rArr(a+3)(a-14/5)=0#
Therefore,

#a+3=0rArra=-3#
OR
#a-14/5=0rArra=14/5#