A 2 digit number is seven times the sum of its digits. The tens digit is 3 more than the units digit. what is the number?
To find the 2-digit number that satisfies these conditions, we can set up an algebraic equation based on the information given.
Let's assume the tens digit is represented by "x" and the units digit is represented by "y".
According to the problem, the number is seven times the sum of its digits. This can be expressed as:
10x + y = 7(x + y)
The problem also states that the tens digit is 3 more than the units digit. So we can write another equation:
x = y + 3
Now we have a system of two equations:
10x + y = 7(x + y)
x = y + 3
To solve this system of equations, we can substitute the value of x from the second equation into the first equation:
10(y + 3) + y = 7(y + 3 + y)
Simplifying, we get:
10y + 30 + y = 7(2y + 3)
Combining like terms:
11y + 30 = 14y + 21
Subtracting 11y from both sides:
30 = 3y + 21
Subtracting 21 from both sides:
9 = 3y
Dividing both sides by 3:
3 = y
Now, substitute the value of y back into the second equation to find the value of x:
x = y + 3 = 3 + 3 = 6
Therefore, the tens digit (x) is 6 and the units digit (y) is 3. So the number is 63.
10 t + u = 7t + 7u
t = u + 3
substituting ... 11 u + 30 = 14 u + 21