SOLUTION: Find a two digit number such that three times the tens digit is 2 less than twice the units digits and twice the number is 20 greater than the number obtained by reversing the di

I won't do yours for you but I'll do one exactly like it, step by step. You
can use it as a model to do yours by.

So instead of your problem, I'll do this one:

The tens digit is t
The units digit is u
The number is 10t+u
The number obtained by reversing the digits is 10u+t

(1)   6t = 4u - 8
(2)   3(10t+u) = (10t+u) + 23

Simplify the equation (2)

      30t + 3u = 10u + t + 23
      
Subtract 10u from both sides

      30t - 7u = t + 23

Subtract t from both sides

(3)   29t - 7u = 23

Now solve equation (1) for t

(1)   6t = 4u - 8

Divide both sides by 6

(4)      t = (4u-8)/6

Substitute in equation (3)

(3)    29t - 7u = 23  
  
29(4u-8)/6 - 7u = 23

Multiply through by 6

 29(4u-8) - 42u = 138 

 116u-232 - 42u = 138

      74u - 232 = 138

Add 232 to both sides

            74u = 370

Divide both sides by 74

              u = 5 

Substitute in equation (4)

(4)      t = (4u-8)/6

         t = [4(5)-8]/6

         t = [20-8]/6

         t = 12/6

         t = 2

So t = tens digit = first digit = 2
and u = units digit = second digit = 5

So the two digit number is 25.

Now use this as a model to do yours by.

Edwin

Let tens and units digits, be T and U, respectively
Then we get:            
                                                            12T - 8U = - 8 ------- Multiplying eq (i) by 4 ------ eq (iii) 
                                                             7T = 28 ------ Subtracting eq (iii) from eq (ii)
                                                             Tens digit, or 

3(4)  -  2U = - 2 ------- Substituting 4 for T in eq (i)
12  -  2U = - 2
-  2U = - 14
Units digit, or