Math Equations II?

Question

Since I WAY messed up last time, I'm retyping it:

In a two-digit number, the units digit is one less than twice the tens digit. If the digits are reversed, the new number is 20 less than twice the original number.

What's the original number? I'm working with substitution equations. So I need to have two equations and combine them. Could you also please show your work so that I can really understand it? Thanks everyone in advance. ^_^

Answer 1

1) the units digit is one less than twice the tens digit: 2x = y + 1
2) If the digits are reversed, the new number is 20 less than twice the original number: 2(x * 10 + y) = y * 10 + x + 20

20x + 2y = 10y + x + 20
eq. 1: 19x = 8y + 20
eq. 2: 2x = y + 1

eq. 2 * 8: 16 x = 8y + 8
eq. 1 : 19 x = 8y + 20

Subtract eq. 2 from eq. 1 to eliminate y

3x = 12
x = 4
y = 7

The answer is 47.

Answer 2

Let the unit digit be x and the tens digit is y
then :x=2*(y)-1 thats mean x=2y - 1
and the other eq. is :x*10+y=2*((y*10)+x)-20
so from eq. 2 :10x + y = 20y + 2x - 20
if we substitute 1 in 2 :
10*(2y -1)+y=20y + 4y-2 -20
20y - 10 +y=24y-22
3y=12 so y=4
x=7
the number is 47 and if you reverse it 74 which is 20 less than twice (47*2=94) the original number.

Answer 3

Let;'s say the tens digit is x and the ones digit is y.
So the number is 10x + y. The first equation is
2x - 1 = y. If you reverse the digits, the new number is
10y + x. So the second equation is
10y + x = 2(10x + y) - 20
You take it from there.

Answer 4

number = xy
reversed = yx
xy = 10x + y
yx = 10y + x

y = 2x -1

yx = 2*xy - 20

10y + x = 2[10x + y] - 20
10y + x = 20x + 2y - 20
8y + 20 = 19x
8/19y + 20/19 = x

y = 2x - 1
y = 2(8/19y + 20/19) - 1
19y = 16y + 40 - 19
3y = 21
y = 7

7 = 2x - 1
8 = 2x
x = 4

xy = 47
yx = 74

Answer 5

42

Answer 6

42! 18 oky 24 set hut-hut-hut HIKE!

Answer 7

answer is 47