Question 1043006
Yes you can plug in any set of values for x, k and h (as long as the denominator isn't zero). So I'm going to plug in x = 0, h = 1 and k = 2.



{{{(2k)/(2x+h) = (k)/(x+h)}}}



{{{(2*2)/(2*0+1) = (2)/(0+1)}}}



{{{(4)/(0+1) = (2)/(0+1)}}}



{{{(4)/(1) = (2)/(1)}}}



{{{4=2}}} Which is a false equation



So the original equation is false. 


Edit: if h = 0 and x was nonzero, then the equation would be true. However, the equation is not true in general. So the answer is still false.