If I understand the question correctly, we are looking for a value #x# such that
#(2+x)/(17+x)=2/5#
We can solve this for #x# by cross-multiplying:
#(2+x)(5)=2(17+x)#
Then distributing:
#=>10+5x=34+2x#
Then moving every thing with #x# to the left, and everything without #x# to the right:
#=>cancel10+5x-color(navy)(cancel(10))=34+2x-color(navy)(10)#
#=>" "5x-color(green)(2x)=24+cancel(2x)-color(green)(cancel(2x))#
#=>" "3x" "=24#
Now we have a simple equation that says "3 times #x# is 24". We solve this by dividing both sides by that 3:
#=>(cancel3x)/cancel3=24/3#
#=>x=8#
So our solution is #x=8#. We can check this by substituting #8# in for #x# in the equation we started with:
#(2+x)/(17+x)=2/5#
#(2+color(blue)8)/(17+color(blue)8)=^?2/5#
#" "10/25" "=^?2/5#
#(2*cancel5)/(5*cancel5)" "=^?2/5#
#" "2/5" "=2/5#
So yes, #x=8# is our solution.