SOLUTION: how to set up: a two digit number is 3 times the sum of its digits. the number is also 45 less than the number formed by reversing the digits of the original number. what is the or
Question 919435: how to set up: a two digit number is 3 times the sum of its digits. the number is also 45 less than the number formed by reversing the digits of the original number. what is the original number? Answer by Theo(13342) (Show Source):
10x + y = 3 * (x + y)
simplify to get:
10x + y = 3x + 3y
subtract 3x from both sides of the equation to get:
7x + y = 3y
subtract y from both sides of the equation to get:
7x = 2y
divide both sides of the equaation by 2 to get:
y = 7x/2
10x + y = 10y + x - 45
subtract x from both sides of the equation to get:
9x + y = 10y - 45
subtract 10y from both sides of the equation to get:
9x - 9y = -45
replace y with 7x/2 to get:
9x - 9 * (7x/2) = -45
simplify to get:
9x - 63x / 2 = -45
multiply both sides of the equation by 2 to get:
18x - 63x = -90
combine like terms to get:
-45x = -90
divide both sides of the equation by -2 to get:
x = 2
y = 7x/2
replace x with 2 to get:
y = (7*2)/2
simplify to get:
y = 7
that's your solution.
x = 2
y = 7
10x + y = 27
10x + y = 3 * (x+y) becomes
10*2 + 7 = 3 * (2 + 7) which becomes:
20 + 7 - 3 * 9 which becomes:
27 = 27 which is true.
10x + y = 10y + x - 45 becomes:
10 * 2 + 7 = 10 * 7 + 2 - 45 which becomes:
20 + 7 = 70 + 2 - 45 which becomes:
27 = 72 - 45 which becomes:
27 = 27 which is true.
everything checks out so the solution is good.
you could also have done a simple logic test up front which would have told you that y had to be equal 7 and x have to be equal to 2.
this is because y = 7x/2
when x = 2, y equal 7.
x had to be between 0 and 9
y had to be between 0 and 9.
x and y both had to be integers.
the only value of x that gave a valid value of y was x = 2.
all others were either out of range of not integers.
