Expert: Sherry Wallin - 7/31/2009

Question
question->
if x/y=6/5,find the value of x^2+y^2/x^2-y^2

aneswar->
x^2+(y^2/x^2)-y^2 = (x^2/y^2 + 1)/(x^2/y^2 - 1)=(6/5)^2+1/(6/5)^2-1=36/25+1/36/25-1=61/25*25/11=61/11


pls explain it  

Answer
Pratap. this is very hard for me to explain because the question is not asked correctly. I know you mean [(x^2/y^2) + 1]/[(x^2/y^2) - 1] for (x^2/y^2 + 1)/(x^2/y^2 - 1) but you are not typing this in correctly, you are violating the order of operations. You are saying
[x^2/(y^2-1)][x^/(y^2+1)] the way you have it typed and it is not true.

What they've done is they have used a rule for squaring fractions and that is that you can either square the numerator and then the denominator or you can square the fraction. An example is:
1^2/2^2 = (1/2)^2
1^1 = 1*1= 1
2^2 =2*2= 4
so (1/2)^2 = 1/4 but so does
(1/2)*(1/2) = [(1*1)/(2*2)]= 1/4

So in your problem they have squared the fraction in the numerator and then added one to that result (6/5)^2 + 1= (6/5)(6/5)+ 1 = 36/25 + 1 = 36/25 + 25/25 = 61/25 and then in the denominator they squared the fraction and subtracted 1 from the result (6/5)^2 -1 =36/25 -25/25 = 11/25 so now you have 61/25 divided by 11/25 = 61/25 * 25/11 = 61/11. I hope this is clear.

Math Prof