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Find a 2 decimal place number between 0 and 1 such that the sum of its digits is 13 and such that when the digits are reversed the number is decreased by 0.45.

I understand there is a standard formula for figuring these out, but I don't really understand why.

I think it's something along the lines of 0.1t +0.01h =13 and 0.01t + 0.1h = 12.55.

But I'm having trouble understanding why this is and applying it to other problems, especially fractions.

The first statement tells us the the sum of the tenths digit (t) and the hundredths digit (h) is 13.  The equation for this is:  t + h = 13

The second statement tells us that reversing the digits decreases the number by .45 this means that .th is .45 greater than .ht.  To get .th, we need to multiply t by .1 and h by .01.  From the second statement, we would get the equation:  .1t + .01h = .1h + .01t + .45

To solve this, I would solve the first equation for one of the variables (I will solve for t) and substitute into the second equation.

t + h = 13

t = 13 - h

.1t + .01h = .1h + .01t + .45

.1(13 - h) + .01h = .1h + .01(13 - h) + .45

1.3 - .1h + .01h = .1h + .13 - .01h + .45

1.3 -.09h = .09h + .58

1.3 - .58 = .09h + .09h

.72 = .18h

h = 4

Now that we have one digit, we can substitute the number into the first equation to find the other digit

t + h = 13

t + 4 = 13

t = 9