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Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: <a href="http://www.bigoo.ws"><img alt="myspace" border="0" src="http://content17.bigoo.ws/content/glitter/cartoon/cartoon_10.gif"></a><p style="margin-top: 0; margin-bottom: 0"><a      Log On


   

Question 30245: myspace



Can you please tell me the answer to this?

%28x%29%2F%28j-k%29-%28x%29%2F%28j%2Bk%29=1
thanks
Answer by sdmmadam@yahoo.com(530) About Me  (Show Source):
You can put this solution on YOUR website!
x/(j-k) - x/(j+k) = 1 ----(1)
Multiplying by (j-k)(j+k) which is the lcm of (j-k)and (j+k)
x(j+k)-x(j-k) = (j-k)(j+k)
x[j+k-j+k] = j^2-k^2 (taking x out on the LHS and applying formula on the RHS)
x(2k) =j^2-k^2
x = (j^2-k^2)/2k
Answer: x = (j^2-k^2)/2k
Verification: x= (j^2-k^2)/2k in (1)
LHS = (j^2-k^2)/2k (j-k) -(j^2-k^2)/2k (j+k)
=(j+k)/2k-(j-k)/2k
=[(j+k)-(j-k)]/2k
=(j+k-j+k)/2k
=(2k)/2k= 1 = RHS