Question 962664
a = the 10's digit
b = the units
ten
10a+b = "the number"
:
Write an equation for each statement, simplify as much as possible
;
A certain number has 2 digits and is equal to the square of the sum of the digits.
10a + b = (a+b)^2
let's leave it at that, see if we can get it down to a single unknown
:
If the number is decreased by 63, the digits are interchanged.
10a + b - 63 = 10b + a
combine on the left
10a - a + b - 10b = 63
9a - 9b = 63
simplify, divide by 9
a - b = 7
a = (b+7)
Back to the 1st equation, replace a with (b+7)
10(b+7) + b = (b+7+b)^2
10b + 70 + b = (2b+7)^2
11b + 70 = 4b^2 + 28b + 49
Combine to form a quadratic equation on the right
0 = 4b^2 + 28b - 11b + 49 - 70
4b^2 + 17b - 21 = 0
you can use the quadratic formula, but this has to factor (integers)
(b-1)(4b+21) = 0
b = 1, the only reasonable solution (has to be an integer)
then
a = 1 + 7
a = 8
:
 Find out the original numbers, 81 is the original number
:
See for yourself if it checks out in the original statements.