Question 164855
Let the tens digit be A, the ones digit be B.
The number is then {{{A*10+B}}}.
"The product of the two-digit number and its tens digit is 54."
1.{{{(A*10+B)*A=54}}}
1.{{{10A^2+AB=54}}}
"the sum of the digits when added to the number gives a result of 36"
2.{{{A+B+(A*10+B)=36}}}
2.{{{11A+2B=36}}}
From eq. 2, 
2.{{{11A+2B=36}}}
{{{2B=36-11A}}}
{{{B=18-(11/2)A}}}
Subsitute this value into eq. 1 and solve for A.
1.{{{10A^2+AB=54}}}
{{{10A^2+A(18-(11/2)A)=54}}}
{{{20A^2+A(36-11A)=108}}}
{{{20A^2+36A-11A^2=108}}}
{{{9A^2+36A-108=0}}}
{{{A^2+4A-12=0}}}
{{{(A+6)(A-2)=0}}}
A negative value for A wouldn't make sense for this problem.
We'll only use the positive value.
{{{A-2=0}}}
{{{A=2}}}
From eq. 2,
{{{B=18-(11/2)A}}}
{{{B=18-(11/2)2}}}
{{{B=18-11}}}
{{{B=7}}}
The two digit number is 27.
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Check your answer.
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"The product of the two-digit number and its tens digit is 54."
{{{27*2=54}}}
{{{54=54}}}
"the sum of the digits when added to the number gives a result of 36"
{{{2+7+27=36}}}
{{{36=36}}}
Both statements are true.
It's a good answer.