Difference of two ninbers is 5 and difference of thier reciprocal is 1/10 find the numbers.
Answer:
X-Y=5
X=Y+5
1/X - 1/Y = 10
1/Y+5 - 1/Y =10
LCM
....
Given that
x-y=5
x=5+y(say-----equation 1)
Now clearly 1/x is less than 1/y because denominator of y is greater than x
therefore
1/y-1/(y+5)=1/10
on solving we get y=5 or -10
when y=5 then x=10
or when y=-10 then x=-5
1/(Y+5) - 1/Y =1/10 <----- you had 10
LCD is y(y+5)
y - (y+5) = y(y+5)/10
-50 = y^2 + 5y
y^2 + 5y + 50 = 0
no real solution.
1/y - 1/(y+5) = 1/10
y = 5
check:
1/5 - 1/10 = 1/10
To find the difference between the two numbers, let's call them X and Y.
According to the given information, the difference between the two numbers is 5: X - Y = 5.
We can rearrange this equation to solve for X: X = Y + 5.
Now, let's consider the second piece of information: the difference between the reciprocals of the two numbers is 1/10.
To express this mathematically, we can write: 1/X - 1/Y = 1/10.
Let's simplify this equation by finding a common denominator for the fractions. The common denominator for X and Y is XY.
Now, multiply both sides of the equation by XY:
Y - X = XY/10.
Since we already found that X = Y + 5, we can substitute this in the equation:
Y - (Y + 5) = (Y + 5)Y/10.
Simplifying further, we have:
-Y - 5 = Y^2 + 5Y/10.
Now, let's continue solving for Y by rearranging the equation:
10(-Y - 5) = Y^2 + 5Y.
Expanding and simplifying, we get:
-10Y - 50 = Y^2 + 5Y.
Now, let's move all terms to one side of the equation:
Y^2 + 5Y + 10Y + 50 = 0.
Combining like terms, we have:
Y^2 + 15Y + 50 = 0.
To solve this quadratic equation, we can either factor it or use the quadratic formula.
Factoring this equation, we get:
(Y + 10)(Y + 5) = 0.
Setting each factor to zero, we find two possible values for Y: Y = -10 and Y = -5.
Now, substitute these values back into the equation X = Y + 5 to find the corresponding values of X:
For Y = -10, X = -10 + 5 = -5.
For Y = -5, X = -5 + 5 = 0.
Therefore, the two numbers are -5 and -10.