Two numbers are reciprocals of each other. One number is 9 times as large as the other. Find the two numbers.

n the number. 1/n the reciprocal

n=9/n
n^2=9
n=3
1/n=1/3

One number is 3 times another. If the sum of their reciprocals is2/9 , find the two numbers.

Well, well, well, aren't these numbers just the "yin" and "yang" of mathematics! Let's call one number "x" and the other number "y". We know their reciprocals are related, so we can write the equation x * y = 1.

Now, we also know that one number is 9 times as large as the other. So we can write another equation, x = 9y.

Substituting the value of x from the second equation into the first equation, we get 9y * y = 1. Simplifying, we have 9y^2 = 1.

To solve for y, we divide both sides by 9: y^2 = 1/9. Taking the square root of both sides, we find that y = 1/3 or y = -1/3.

So the two numbers are either 1/3 and 9/1 (which simplifies to 9), or -1/3 and -9.

To solve this problem, let's start by assigning variables to the two numbers. Let's say the first number is x and the second number is y.

According to the given information, the two numbers are reciprocals of each other. This means that their product is equal to 1:

xy = 1 ----(1)

The problem also states that one number is 9 times as large as the other. This can be expressed as:

x = 9y ----(2)

Now we have two equations (equations 1 and 2) with two variables (x and y). We can solve this system of equations to find the values of x and y.

Substitute equation (2) into equation (1) to eliminate x:
(9y)y = 1

Simplify the equation:
9y^2 = 1

Divide both sides of the equation by 9:
y^2 = 1/9

Take the square root of both sides:
y = 1/3

Now substitute the value of y back into equation (2) to find x:
x = 9(1/3)
x = 3

Therefore, the two numbers are 3 and 1/3.