```Question 448333
I'll do the first one to get you started. If you need more help, then repost.

# 3

Let u = units digit and t = tens digit

So the original number is then {{{10t+u}}}. The reversed digit number is {{{10u+t}}} (swap u and t). Ex: if t = 5 and u = 9, then the original number is 59 and the swapped number is 95.

Since "The units digit is 3 times more the tens digit", we know that {{{u=3t}}} (ie multiply the tens digit t by 3 to get the units digit u).

Now "When the digits are reversed, the new number is 54 more than the original", this means that {{{10u+t=10t+u+54}}}

{{{10(3t)+t=10t+3t+54}}} Plug in {{{u=3t}}} (ie replace EVERY "u", without quotes, with "3t", without quotes)

{{{30t+t=10t+3t+54}}} Multiply

{{{31t=10t+3t+54}}} Combine like terms on the left side.

{{{31t=54+13t}}} Combine like terms on the right side.

{{{31t-13t=54}}} Subtract {{{13t}}} from both sides.

{{{18t=54}}} Combine like terms on the left side.

{{{t=(54)/(18)}}} Divide both sides by {{{18}}} to isolate {{{t}}}.

{{{t=3}}} Reduce.

So the tens digit is 3. Because we know that {{{u=3t}}}, we can plug this into that equation to get {{{u=3(3)=9}}}. So the units digit is 9

So the original number is 39. The new swapped number is 93. Notice how 93 is 54 more than 39 (ie 39+54 = 93)

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# 4

Use the same basic strategy to solve this problem. Let me know if you still need help.```