the sum of the digits of a two-digit number is 12. when the digits are reversed, the new number is 18 less than the original number. check your answer. HELP~

Question

mr.math · Answer

If the number has the form $a_0a_1$, then we can make the system:
$$a_0+a_1=12$$
$$10a_0+a_1=10a_1+a_0+18$$

pokemon23 · Answer

hola turningtest goodnight people

unklerhaukus · Answer

57,75

anonymous · Answer

I'm lost :c

turingtest · Answer

...this exhibits my love/hate relationship with Diophantine equations. Nice formulation Mr. Math :)

mr.math · Answer

Let me explain. You have a number of two digits, call it $a_0a_1$, where $a_0$ is the 10's digit and $a_1$ is the 1's digit. Now we're given that the sum of the two digits is 12. We can write that as $a_0+a_1=12$. Following so far?

anonymous · Answer

ohh yeah, ohkay keep going..

mr.math · Answer

Good. The second information we are given is that when reversing the digits [that's writing it as $a_1a_0$], the new number is 18 less than the original number. First you need to notice that if the original number is x, we can write $x=10a_0+a_1$. Call the new number y, then $y=10a_1+a_0$. Hence the second equation of the system is:
$$10a_0+a_1=10a_1+a_0+18.$$

mr.math · Answer

All you're left to do here is to solve the system for $a_0$ and $a_1$.

anonymous · Answer

Can you do the steps and explain to me please? I seem to learn better by looking at the steps done.

turingtest · Answer

$$a_0+a_1=12\to a_0=12-a_1$$$$10a_0+a_1=10a_1+a_0+18\to9a_0-9a_1=18$$multiply first eqn by 9$$9a_0+9a_1=108$$$$9a_0-9a_1=18$$add these together...$$18a_0=126\to a_0=7$$now we can find the other number with the formula for a_1 on the first line$$a_0+a_1=12\to a_1=12-a_0=12-7=5$$hence $$a_0=7$$$$a_1=5$$so the numbers are$$57,75$$as UnkleRhaukus said

anonymous · Answer

the numbers are 57, and 75 but which is the original?

unklerhaukus · Answer

the original is 75