The sum of two numbers is 81, one of the numbers is 9 more than the other.What are the numbers?

please help..thanks a lot

i think of 3 positive numbers. the greatest number is twice a perfect square. the least of them is 1 less than half the greatest number. the average of the 3 numbers is 12.

To find the two numbers, let's assign variables to represent them. Let's call one number "x" and the other number "y".

According to the problem, the sum of the two numbers is 81. So, we can write the equation:
x + y = 81

The problem also states that one of the numbers is 9 more than the other. We can write this relationship as:
x = y + 9

Now we have a system of two equations. We can solve this system using substitution or elimination method.

Since we have x = y + 9, we can substitute this value into the first equation.

Substituting x + y = 81:
(y + 9) + y = 81
2y + 9 = 81
2y = 81 - 9
2y = 72
y = 72/2
y = 36

Now that we have y = 36, we can substitute this value back into x = y + 9 to find x.

Substituting y = 36:
x = 36 + 9
x = 45

So, the two numbers are 36 and 45.

To find the two numbers, we can set up a system of equations based on the given information. Let's say one of the numbers is "x", then the other number would be "x + 9" because it is 9 more than the first number.

So, we have the equation:
x + (x + 9) = 81

Now we can simplify this equation by combining like terms:
2x + 9 = 81

Next, we isolate the variable by moving 9 to the other side of the equation:
2x = 81 - 9
2x = 72

Finally, we solve for x by dividing both sides of the equation by 2:
x = 72 / 2
x = 36

So one of the numbers is 36, and to find the other number we can substitute the value of x into our equation:
x + 9 = 36 + 9 = 45

Therefore, the two numbers are 36 and 45.

x + (x + 9) = 81

2x = 72

x = 36

x + 9 = 45