Question 226573
Let x = original number and y = reversed number


Any two digit number is of the form {{{x=10t+u}}}. Ex: 21=10*2+1 (think of t is the tens digit and u as the units/ones digit)



So {{{x=10t+u}}} and {{{y=10u+t}}} (just swap t and u)



The statement "if a certain two digit number is divided by the sum of its digits, the quotient is 3" translates to {{{x/(t+u)=3}}} (because t+u is the sum of the digits). Plug in {{{x=10t+u}}} to get {{{(10t+u)/(t+u)=3}}}. So this is our first equation.



"if the digits are reversed, the new number is 9 less then three times the original number" translates to {{{y=3x-9}}}. Plug in {{{x=10t+u}}} and {{{y=10u+t}}} to get {{{10u+t=3(10t+u)-9}}}. This is the second equation.



So you now have the system {{{system((10t+u)/(t+u)=3,10u+t=3(10t+u)-9)}}}



Because you have two equations with two unknowns, you can find the unique solution which means you can find t and u. I'll let you do that. Let me know if you still need help.