The sum of two numbers is 12. The sum of their reciprocal is 3/8? what are the numbers?
a+b = 12
1/a + 1/b = 3/8
4+8 = 12
To find the two numbers, let's assume the first number is x and the second number is y.
The problem states that the sum of the two numbers is 12, so we have the equation:
x + y = 12 -- (Equation 1)
The problem also states that the sum of their reciprocals is 3/8. The reciprocal of a number is 1 divided by that number. So, the equation for the sum of their reciprocals is:
1/x + 1/y = 3/8 -- (Equation 2)
To solve this system of equations, we can use substitution or elimination:
Let's solve using substitution:
From Equation 1, we can rewrite it as:
y = 12 - x
Now, substitute this expression for y in Equation 2:
1/x + 1/(12 - x) = 3/8 -- (Equation 3)
To simplify this equation, we can multiply everything by 8x(12-x) to eliminate the fractions:
8(12 - x) + 8x = 3x(12 - x)
Simplify further:
96 - 8x + 8x = 36x - 3x^2
Combining like terms:
96 = 36x - 3x^2
Rearrange the terms:
3x^2 - 36x + 96 = 0
Now, let's solve this quadratic equation. We can either factor it or use the quadratic formula. Let's use factoring:
3x^2 - 36x + 96 = 0
Divide through by 3 to simplify:
x^2 - 12x + 32 = 0
Now we need to factor this equation. We are looking for two numbers whose sum is -12 and whose product is 32.
The factors are: -4 and -8
So, re-writing Equation 3:
(x - 4)(x - 8) = 0
This means that either (x - 4) = 0 or (x - 8) = 0.
Therefore, we have two possible values for x:
x = 4 or x = 8
Now, substitute these values for x in Equation 1 to find the corresponding values for y:
For x = 4:
y = 12 - x
y = 12 - 4
y = 8
So, one pair of numbers is x = 4 and y = 8.
Similarly, for x = 8:
y = 12 - x
y = 12 - 8
y = 4
So, the other pair of numbers is x = 8 and y = 4.
Thus, the two pairs of numbers that satisfy the given conditions are: (4, 8) and (8, 4).