The larger of two numbers is twice the smaller number. The sum of their reciprocals is 1. Find the larger number.
1/x + 1/2x = 1
3x = 2x^2
x = 3/2
2x = 3
check:
1/(3/2) + 1/(3) = 2/3 + 1/3 = 1
Let the larger number be y
hence the smallest number is 0.5y
their sum = y + 0.5y
their recipocal = 1/(y+0.5y)
1/(y+0.5y)=1
2 = 2y + y
2 = 3y
y=2/3
hence the larger number is 2/3
Let the larger number be y
hence the smallest number is 0.5y
their sum = y + 0.5y
their reciprocal = 1/(y+0.5y)
1/(y+0.5y)=1
2 = 2y + y
2 = 3y
y=2/3
hence the larger number is 2/3
To solve this problem, let's denote the smaller number as "x" and the larger number as "2x". We are given two conditions:
1. The larger number is twice the smaller number: 2x = 2 * x = 2x.
2. The sum of their reciprocals is 1: 1/x + 1/2x = 1.
Now, let's solve the equation:
1/x + 1/(2x) = 1
To add these fractions, we need to find a common denominator, which is 2x:
(2 + 1) / (2x) = 1
(3/2) / (2x) = 1
To remove the fraction, we multiply both sides by (2x):
(3/2) = 2x
Now, we can solve for "x":
(3/2) = 2x
Multiplying both sides by (2/3):
(3/2) * (2/3) = 2x * (2/3)
1 = (4/3)x
To solve for "x", we divide both sides by (4/3):
1 / (4/3) = (4/3)x / (4/3)
(3/4) = x
So, the smaller number (x) is 3/4.
Now, let's find the larger number by substituting x back into the equation 2x:
2 * (3/4) = 6/4 = 3/2
Therefore, the larger number is 3/2.