How do you solve #3/(5-x)+2/(4-x) = 8/(x+2)# ?

The quickest way to find at least one solution is to "guess":

Note that #3/(5-x)# will divide evenly if #5-x=3#, i.e. when #x=color(blue)(2)#

Then we find:

#3/(5-color(blue)(2)) + 2/(4-color(blue)(2)) = 3/3+2/2 = 1+1 = 2 = 8/4 = 8/(color(blue)(2)+2)#

#color(white)()#
Are there any other solutions?

Given:

#3/(5-x)+2/(4-x) = 8/(x+2)#

Subtract #3/(5-x)+2/(4-x)# from both sides to get:

#0 = 8/(x+2)-3/(5-x)-2/(4-x)#

#color(white)(0) = 8/(x+2)+3/(x-5)+2/(x-4)#

Multiply through by #(x+2)(x-5)(x-4)# to get:

#0 = 8(x-5)(x-4)+3(x+2)(x-4)+2(x+2)(x-5)#

#color(white)(0) = 8(x^2-9x+20)+3(x^2-2x-8)+2(x^2-3x-10)#

#color(white)(0) = (8x^2-72x+160)+(3x^2-6x-24)+(2x^2-6x-20)#

#color(white)(0) = 13x^2-84x+116#

We can solve this quadratic by any method we like, but we already know that #x=2# is a zero of this quadratic, so #(x-2)# is a factor:

#13x^2-84x+116 = (x-2)(13x-58)#

So the other solution is #x = 58/13#

#3/(5-x)+2/(4-x)=8/(x+2)# can be simplified as

#(3(4-x)+2(5-x))/((5-x)(4-x))=8/(x+2)# or

#(12-3x+10-2x)/(5(4-x)-x(4-x))=8/(x+2)# or

#(22-5x)/(20-5x-4x+x^2)=8/(x+2)# or

#(22-5x)/(20-9x+x^2)=8/(x+2)# or

#8(20-9x+x^2)=(x+2)(22-5x)#

#160-72x+8x^2=x(22-5x)+2(22-5x)# or

#160-72x+8x^2=22x-5x^2+44-10x# or

#8x^2+5x^2-72x-22x+10x+160-44=0# or

#13x^2-84x+116=0# or

#13x^2-26x-58x+116=0# or

#13x(x-2)-58(x-2)=0# or

#(13x-58)(x-2)=0#

Hence #x=2# or #x=58/13#