The ten digit of a two-digit positive integer is 2 more than three times the ones digit. If the digits are interchanged, the new number is 13 less than half the given number. Find the given integer.
Let x = tens-place digit and y = ones-place digit, then 10x + y is the number. Explain how you did this.
2006-11-12 02:56:56 · 2 answers · asked by xolifes2sh0rtox 1 in Science & Mathematics ➔ Mathematics
T=2+3O
T-3O=2
!0O+T=1/2(10T+O)-13
multiplying by 2
20O+2T=10T+O-26
19O-8T=-26
-24O+8T=16
adding
-5O=-10
dividing by 5
O=2
sub in (1)
T-6=2
T=8
original No is 82
2006-11-12 03:04:14 · answer #1 · answered by raj 7 · 0⤊ 0⤋
Let the number be 10x + y. (as u mentioned)
x = 3y+2
So you have a new equation.
10(3y+2) +y = 31y +20
Switch the numbers, 10y + x
= 10y + 3y + 2
= 13y + 2
So, 0.5 (31y+20) - 13 = 13y + 2
15.5y - 3 = 13y + 2
2.5y = 5
y= 2
x = 3(2) + 2
= 8.
So the digit is 82.
Hope this helps=)
2006-11-12 11:17:29 · answer #2 · answered by luv_phy 3 · 0⤊ 0⤋