Question 126664
You know that distance = rate X time or {{{d=rt}}}, or, since we are trying to determine a rate: {{{r=d/t}}}.


For both the upstream and downstream trips we know the distance is 6 miles and the times, 4 hours and 3 hours respectively.


For the upstream trip, the true rate of speed of the boat is the boat's rate of speed in still water MINUS the rate of the current.  For the downstream trip, the true rate is the still water rate PLUS the rate of the current.


Upstream equation:  {{{r[s]-r[c]=6/4=3/2}}}, where {{{r[s]}}} is the rate in still water and {{{r[c]}}} is the rate of the current.


Downstream equation:  {{{r[s]+r[c]=6/3=2}}}


{{{r[s]-r[c]=3/2}}}
{{{r[s]+r[c]=2}}}


Now add the equations, term-by-term:
{{{2r[s]+0r[c]=7/2}}}


And solve:
{{{2r[s]=7/2}}}
{{{r[s]=7/4}}}


So the rate in still water is {{{7/4}}} mph.


Check:
To go 6 miles in 4 hours, the true rate must have been {{{3/2}}} mph.  To go 6 miles in 3 hours, the true rate must have been {{{2}}} mph.


{{{7/4-r[c]=3/2}}} means that {{{r[c]=1/4}}}.   {{{7/4+1/4=8/4=2}}}, answer checks.


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Let the 10s digit be x and the 1s digit be y.  The sum of the digits is 11, so:

{{{x+y=11}}}


And the original number is n, where {{{n=10x + y}}}


If you reverse the digits, the new number which is 45 more than the old number would be given by:


{{{n+45=10y + x}}}, but this can be written:  {{{n=10y+x-45}}}.


Now we have two expressions that are equal to n, so we can set these two expressions equal to each other:


{{{10x+y=10y+x-45}}}
{{{10x-x-10y+y=-45}}}
{{{9x-9y=-45}}}
{{{x-y=-5}}}


From our first equation: {{{x+y=11}}}, we can get {{{y=11-x}}} and we can substitute this expression for y into the previous result:

{{{x-(11-x)=-5}}}
{{{2x-11=-5}}}
{{{2x=-5+11}}}
{{{2x=6}}}
{{{x=3}}}


Now substituting this value for x into the first equation:
{{{3+y=11}}}
{{{y=8}}}


So the original number is 38.


Check:

3 + 8 = 11.

83 - 38 = 45.  Answer checks.