<PRE><B>Question 27579</B>
): Please help me solve this.The sum of two numbers is twenty-one. Three times the smaller number is two less than twice the larger number. Find the two numbers.Thank you
Let the required numbers be a and b with a&lt;b (that is a smaller than b)
Given  a + b = 21 ----(1) (the sum)
3 times the smaller number is 2 less than twice the larger number.
That is 3a is 2 less than 2b
That is 2b - 3a = 2 ----(2)
From (1), we have a = (21-b) ----(*)
Putting * in (2), we have 
  2b - 3(21-b) = 2
  2b -63 + 3b = 2
(2b+3b) = 2 +63 (taking -63 to RHS, change side then change sign)
   5b = 65
    b = 65/5 = 13
Put b = 13 in (*)
 a + b = 21
 a + 13 = 21 which implies a = 21-13 = 8
Therefore the numbers are 8 and 13.
Verification:
3 times the smaller number should be  2 less than twice the larger number.
That is 3a should be less than 2b by 2
And  3X8=24  is of course 2 less than 2X13 = 26 
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