one positive integer is 3 more than another. the difference of their reciprocals is 1/6. find the twonumbers.
1/a - 1/(a+3) = 1/6
a = 3
check:
1/3 - 1/6 = 1/6
To solve this problem, let's set up the equations based on the given information.
Let's say the first number is x, and the second number is y.
According to the problem, one positive integer (let's assume it's x) is 3 more than another (which is y). So we can write the equation:
x = y + 3 ----(Equation 1)
The problem also mentions that the difference of their reciprocals is 1/6. So we can write the equation:
1/x - 1/y = 1/6 ----(Equation 2)
To solve this system of equations, we can use substitution method.
Substitute the value of x from Equation 1 into Equation 2:
1/(y + 3) - 1/y = 1/6
To eliminate the fractions, we can multiply through by the least common multiple (LCM) of the denominators, which is 6y(y + 3):
6y - 6(y +3) = y(y + 3)
Simplify the equation:
6y - 6y - 18 = y^2 + 3y
Combine like terms:
-y^2 - 3y - 18 = 0
Rearrange the equation:
y^2 + 3y + 18 = 0
Now we have a quadratic equation. We can solve it by factoring, completing the square, or using the quadratic formula. However, this quadratic equation has no real solutions because the discriminant (b^2 - 4ac) is negative.
Therefore, there are no real positive integers that satisfy the given conditions.
However, if you allowed non-integer solutions, you could use the quadratic formula to find the values of y.