One Number Is 3 Times Another Number. If The Sum Of The Reciprocals Of The Numbers Is 1/6, What Are The Two Numbers?

To solve this problem, let's first represent the two numbers as variables. Let's call the first number "x" and the second number "y".

We are given that one number is 3 times the other number, so we can set up an equation:

x = 3y

Next, we are told that the sum of the reciprocals of the numbers is 1/6. The reciprocal of a number "a" is 1/a. So, we can set up the equation:

1/x + 1/y = 1/6

Now, we have a system of two equations with two variables. We can solve this system to find the values of x and y.

Substituting x = 3y into the second equation, we get:

1/(3y) + 1/y = 1/6

To add fractions with different denominators, we need to find a common denominator. In this case, the common denominator is 3y. So, we can rewrite the equation as:

(1 + 3)/(3y) = 1/6

Simplifying the left side of the equation, we get:

4/(3y) = 1/6

To solve for y, we can cross-multiply:

4 * 6 = 3y

24 = 3y

Dividing both sides by 3, we get:

y = 8

Now that we have the value of y, we can substitute it back into the equation x = 3y to find x:

x = 3 * 8
x = 24

Therefore, the two numbers are 24 and 8.

Math is NOT my field, but in general:

The reciprocal of a fraction is obtained by interchanging the numerator and the denominator, i.e. by inverting the fraction.

The reciprocal of 1/6 is 6/1. Looking at the 6, 2 X 3 = 6. But when it comes to fractions, I'm lost!

Sra

Please put your subject in the "School Subject" space.

x = 3y

1/x + 1/y = 1/6

Substitute 3y for x in the second equation and solve for y. Put that value in the first equation to find x. Check by putting both values into the second equation.