The difference of two numbers is 47. The larger number is three less than six times the smaller number. Find the numbers.

L = S + 47

L = 6S - 3

Substitute S+47 for L in the second equation and solve for L. Insert that value into the first equation to solve for S. Check by putting both values into the second equation.

S+47=6S-3

+3 +3
S+50=6S
-S -S
50= 5S
10=S
L=S+47
L=(10)+47
L=57

To find the two numbers, let's denote the larger number as "x" and the smaller number as "y".

From the given information, we know that the difference between the two numbers is 47, so we can write the equation:
x - y = 47 ----(1)

We are also told that the larger number is three less than six times the smaller number, which can be written as:
x = 6y - 3 ----(2)

Now we can solve the system of equations by substituting equation (2) into equation (1) to eliminate x:

6y - 3 - y = 47

Combining like terms:
5y - 3 = 47

Adding 3 to both sides:
5y = 50

Dividing both sides by 5:
y = 10

Now substitute the value of y back into equation (2) to find x:

x = 6(10) - 3
x = 60 - 3
x = 57

Therefore, the smaller number is 10 and the larger number is 57.