Digit Problem

[onmyown]

Hi, I am having trouble solving this problem. I have made at least 4-5 attempts on it by now. Can someone please help me out?

Here is the problem: A number consists of two digits. 13 more than 7 times the units digits is 3 times the tens digit. If the digits are reversed, 2 times the new number is 34 less than the original number. Find the number.

Here is one my of attempts:
- Translated each statement into equations:
"13 more than 7 times the units digits is 3 times the tens digit." --> 13+7u=3t
"If the digits are reversed, 2 times the new number is 34 less than the original number" --> 2(10u+t)=10t+u-34
- Made all variables on one side of each equation, and made a system of equations:
3t-7u=13
19u-8t= (-34)
- Solved the system of equations, to solve for u:
3 * (8t+19u = -34) ------> 24t+57u=-102
8 * (3t - 7u = 13) ------> - 24t - 56u=104
113u/113 = -206/113
Therefore, u= -1.8
- Substituted value of u into other equation, to solve for t:
3t-7u=13
3t-7(-1.8)=13
3t+12.6-12.6=13-12.6
3t/3=0.4/3

Therefore, t=0.13

I know all that is wrong. The answer is supposedly 92. I'd appreciate any help, thanks

[stapel_eliz]

A number consists of two digits. 13 more than 7 times the units digits is 3 times the tens digit. If the digits are reversed, 2 times the new number is 34 less than the original number. Find the number.

Here is one my of attempts:

Thank you for showing your work so nicely!

I get the same set-up you do:

3t-7u=13
19u-8t= (-34)

...but you appear to have switched a sign in the second equation:

- Solved the system of equations, to solve for u:
3 * (+8t+19u = -34) ------> 24t+57u=-102

Correcting, you would get:

. . . . ..3t - ..7u = .13
. . . . .-8t + 19u = -34

. . . . .8[..3t -..7u = .13]
. . . . .3[-8t + 19u = -34]

. . . . ..24t - 56u = .104
. . . . .-24t + 57u = -102
. . . . . .0t + ..1u = ...2

As one would expect for this sort of exercise, yes, the values for the variables will need to be whole numbers.