find two numbers whose sum is 25 and the sum of whose reciprocals is one sixth
I tried to finish drwls solution and verify the answers, but they would not check out.
After my second cup of coffee, I realized that my good collegue made what we used to call "wishful thinking solutions" error.
the factors would be (x-15)(x-10)
and we get symmetric solutions, that is, when x=15, y=10 and when x=10 y = 15
Solve this pair of equations:
x + y = 25
1/x + 1/y = 1/6
Rewrite as
1/x + 1/(25-x) = 1/6
[(25-x) + x]/[x(25-x)] = 1/6
150 = x(25-x)
x^2 -25x + 150 = 0
(x-30)(x-5) = 0
One more step
Correct, my mistake.
x+25x+150=0
To find two numbers whose sum is 25, let's assume one number as 'x' and the other as '25 - x'. This is because if we add two numbers and the sum is 25, subtracting one from the other will also give us the same sum.
The sum of their reciprocals is one sixth. We know that a reciprocal of a number is obtained by dividing 1 by that number. So, the reciprocal of 'x' is 1/x, and the reciprocal of '25 - x' is 1/(25 - x).
According to the problem, the sum of their reciprocals is one sixth. Therefore, we can write the equation:
1/x + 1/(25 - x) = 1/6
To solve this equation, we need to find the values of 'x' that satisfy it.
Multiplying through by 6x(25 - x) to eliminate the denominators gives us:
6(25 - x) + 6x = x(25 - x)
Expanding the equation further, we have:
150 - 6x + 6x = 25x - x^2
Rearranging the terms, we get:
x^2 - 25x + 150 = 0
We have a quadratic equation now. We can solve it using factoring, completing the square, or the quadratic formula. Let's use factoring in this case.
The quadratic equation factors into:
(x - 10)(x - 15) = 0
This implies that x can be either 10 or 15.
So, the two numbers whose sum is 25 are 10 and 15.