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:
Find a two digit number such that six times the tens digit is 8 less than
four times the units digits, and three times the number is 23 greater than the
number obtained by reversing the digits.
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