if the reciprocal of the smaller of two consecutive integers is subtracted from three times the reciprocal of the larger, the result is 3/10. find
the integers
The question led me to a quadratic equation: let 1st nos be n,and 2nd be n+1 3/n+1 - 1/n=3/10,collect lcm:
3n-1(n+1)/n(n+1)=3/10. 10(2n-1)=3(n*n)+3n,20n-10=3(n*n)+3n,take like terms,find n,choose the integer as n's value.
smaller number --- x
next consecutive number ---- x+1
3(1/(x+1)) - 1/x = 3/10
multiply each term by 10x(x+1), the LCM
30x - 10(x+1) = 3x(x+1)
30x - 10x - 10 = 3x^2 + 3x
3x^2 - 17x + 10 = 0
(x - 5)(3x - 2) = 0
x = 5 or x = 2/3
but our number was to be an integer, so
x = 5
the integers are 5 and 6
check:
3(1/6) - 1/5
= 1/2 - 1/5
= 3/10
My answer is correct
To solve this problem, let's represent the smaller integer as x and the larger integer as (x+1).
According to the problem, we are given the following equation:
3 * (1 / (x+1)) - 1 / x = 3/10
To simplify the equation, we can find a common denominator:
3 * (x / (x * (x+1))) - (x+1) / (x * (x+1)) = 3/10
Now, we can combine the terms:
(3x - (x+1)) / (x * (x+1)) = 3/10
Expanding the numerator:
(3x - x - 1) / (x * (x+1)) = 3/10
(2x - 1) / (x * (x+1)) = 3/10
To eliminate the fractions, we can cross-multiply:
10(2x - 1) = 3(x * (x+1))
Distributing:
20x - 10 = 3x^2 + 3x
Rearranging the equation to form a quadratic equation:
3x^2 + 3x - 20x + 10 = 0
3x^2 - 17x + 10 = 0
Now, we can solve this quadratic equation using factoring, completing the square, or the quadratic formula. Let's use factoring:
(3x - 2)(x - 5) = 0
Setting each factor to zero:
3x - 2 = 0 or x - 5 = 0
Solving for x:
3x = 2 or x = 5
x = 2/3 or x = 5
Since x represents an integer, the smaller integer can only be x = 5.
Therefore, the two consecutive integers are 5 and 6.