The sum of the digits of a certain two-digit number is 7. Reversing its digits increases the number by 9.
What is the number?
TIA
To solve this problem, we can use algebraic representation and equations. Let's use the following variables to represent the digits of the two-digit number:
Let the tens digit be "x"
Let the units digit be "y"
According to the given information, the sum of the digits is 7. Hence, we have the equation:
x + y = 7 -- Equation 1
It is also given that reversing the digits of the number increases the number by 9. Since the original number is a two-digit number, reversing its digits means swapping the tens and units digits. So, the reversed number is "10y + x".
The original number is "10x + y" and reversing it increases it by 9:
10y + x = 10x + y + 9 -- Equation 2
Simplifying equation 2, we get:
9x - 9y = 9
Now, we have a system of equations:
x + y = 7 -- Equation 1
9x - 9y = 9 -- Equation 3
To solve the system of equations, we can use the method of substitution. Rearrange equation 1 to obtain x in terms of y:
x = 7 - y
Substitute this value of x into equation 3:
9(7 - y) - 9y = 9
Simplify and solve for y:
63 - 9y - 9y = 9
-18y = 9 - 63
-18y = -54
y = -54 / -18
y = 3
Substitute the value of y back into equation 1 to find x:
x + 3 = 7
x = 7 - 3
x = 4
Therefore, the tens digit (x) is 4 and the units digit (y) is 3. The two-digit number is 43.
[ab] = 10 a + b
a+b = 7
[ba] = 10 b + a = 9 + 10a + b
two equations
a + b = 7
and
10 b + a = 9 + 10 a + b
9 b = 9 + 9 a
b = 1 + a
combine
a + b - b = 7 - 1 - a
2 a = 6
a = 3
then b = 4
so answer is
34
check
34 + 9 = 43 sure enough