What number must be added to both the numerator and denominator of #2/17# to produce a fraction equivalent to #2/5#?

1 Answer
Dec 5, 2016

The answer is #8#.

Explanation:

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.