If x/y = y/x + 3/2 find the value of x^2/y^2 +y^2/x^2
Answers
X/Y = 3/2 So X = 3Y/2
(X^2 + Y^2)/(X^2 - Y^2)
Now replace the X value from first equation
= [ (3Y/2)^2 + Y^2] / [ ((3Y/2)^2 - Y^2]
= [ (9Y^2/4) + Y^2] / [ (9Y^2/4) - Y^2]
= [ (9Y^2 + 4Y^2)/4] / [ (9Y^2 - 4Y^2)/4]
= (9Y^2 + 4Y^2) / (9Y^2 - 4Y^2)
= 13Y^2 / 5Y^2
= 13/5
If you have to solve this in any competition exam then just simply replace any value in X & Y which satisfy the equation X/Y = 3/2.
Example: X = 3 & Y = 2 and X and Y should not equal to 0.
= (X^2 + Y^2)/(X^2 - Y^2)
= 3^2 + 2^2 / 3^2 - 2^2
= 9 + 4 / 9 - 4
= 13/5
This will save more time. I hope you understand.
Answer:
Step-by-step explanation:
X/Y = 3/2 So X = 3Y/2
(X^2 + Y^2)/(X^2 - Y^2)
Now replace the X value from first equation
= [ (3Y/2)^2 + Y^2] / [ ((3Y/2)^2 - Y^2]
= [ (9Y^2/4) + Y^2] / [ (9Y^2/4) - Y^2]
= [ (9Y^2 + 4Y^2)/4] / [ (9Y^2 - 4Y^2)/4]
= (9Y^2 + 4Y^2) / (9Y^2 - 4Y^2)
= 13Y^2 / 5Y^2
= 13/5
IF IT WAS HELPFUL KINDLY MARK ME AS BRAINLIEST.