SOLUTION: The sum of the digits of a three-digit number is 9. Four times the tens digit minus 3 times the units digit is 6. If 2 times the units digit is added to twice the hundreds digit, t

You can do it woth or without algebra.  I'll do it both ways:

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Without algebra:

If 2 times the units digit is added to twice the hundreds digit, 
the result is 12.

That means that 

If 1 times the units digit is added to 1 times the hundreds digit,
the result is 6.

Since the sum of all three digits is 9, and the sum of the units and 
the hundreds digits is 6, the tens digit has to be 3.

So, four times the tens digit is 12.

We are told that four times the tens digit minus (3 times the units digit)
is 6.

So 12 minus (3 times the units digit) is 6.

So (3 times the units digit) must be 6.

Therefore the units digit is 2.

And since the sum of all three digits is 9,  and the tens digit is 3
and the units digit 2, the hundreds digit must be 4.

So the number is 432.
 
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By algebra

H + T + U = 9
4T - 3U = 6
2U + 2H = 12

Solve that and get H=4, T=3 and U=2

Edwin