|
Question 890107: A two-digit number is thrice as large as the sum of its digits,and the square of that sum is equal to the trebled required number.Find the number?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website!
Let a = the 10's digit
let b = the units:
then
10a+b = the number
:
Write an equation for each statement, simplify as much as you can:
:
A two-digit number is thrice as large as the sum of its digits,
10a+ b = 3(a+ b)
10a + b = 3a + 3b
10a - 3a = 3b - b
7a = 2b
divide both sides by 2
b - 3.5a
:
and the square of that sum is equal to the trebled required number.
(a+b)^2 = 3(10a+b)
replace b with 3.5a
(a+3.5a))^2 = 3(10a + 3,5a)
(4.5a)^2 = 3(13.5a)
20.25a^2 = 40.5a
divide both sides by a
20.25a = 40.5
a = 40.5/20.25
a = 2
find b
b = 3.5(2)
b = 7
:
Find the number? It's 27
;
:
Check this in the 2nd statement
the square of that sum is equal to the trebled required number"
9^2 = 3(27)
|
|
|
| |
