Question 1166205
{{{100h+10t+u}}}, the number


{{{h+t+u=11,h-u=t-1}}}-----not yet finished analyzing the whole description.

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{{{h=u+t-1}}}
{{{u+t-1=h}}}
{{{t+u-h=1}}}


Expressing the system of those two equations again as:

{{{system(h+t+u=11,h-t-u=-1)}}}
ADD corresponding members.
{{{2h=10}}}
{{{highlight_green(h=5)}}}-----------the hundreds digit appears to be 5.  


Looking at the original two-equation system again sub for h
{{{5+t+u=11,5-u=t-1}}}
-
{{{highlight_green(t+u=6)}}}



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If the hundreds and tens digits are interchanged, the number is reduced by 90. 
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{{{100t+10h+u+90=100h+10t+u}}}


{{{100t+50+u+90=500+10t+u}}}

{{{100t+140=500+10t}}}

{{{90t=500-140}}}

{{{9t=50-14}}}

{{{9t=36}}}

{{{highlight_green(t=4)}}}


.

.
{{{highlight_green(u=2)}}}


Original Number,  {{{highlight(highlight(542))}}}