Question 970051
{{{10t+u}}} the two digit number.


{{{10t+u+10t+u=(10t+u)}}}-----?  Product with WHAT?  With itself?



{{{10t+u+10t+u=(10t+u)^2}}}
{{{20t+2u=100t^2+20tu+u^2}}}
Not sure that this is even within College Algebra, seeing a {{{tu}}} term, and only one equation but two variables.  You expect only t and u are digits.  One way although lengthy, is solve for t and test all values of u from the digits 0 through 9.  With each of those u values, finish solving for t, which also must be a digit.


Just checking your expected "22",
10*2+2+10*2+2=22*22
20+2+20+2=2*2*11*11
44=4*121
NO GOOD!


Better, <b>what is the exact problem description and question, word-for-word?</b>



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New Description:  Sum of the digits is equal to the product of the digits, for an unknown two digit number.


t and u for the Tens place and the Ones place.
{{{t+u=t*u}}}
Just one equation but two unknown variables.
You know that u and t must be from the set of digits.  You can start with u at 0 and cycle through to u at 9 and solve t for each case until something works  (meaning t is found as a digit).  You first find that u=0 gives you nothing.


{{{t-tu=-u}}}
{{{t(1-u)=-u}}}
{{{t=-u/(1-u)}}}
{{{highlight_green(t=u/(u-1))}}}


u________________t
0________________0
1______________meaningless by undefined
2_______________2
3_______________{{{3/2}}}
4_______________{{{4/3}}}
5_______________{{{5/4}}}
Following the pattern you see the rest will not work.


Only one of those combinations gave a digit for t.
That is the number 22.