Question 938084
Let {{{x}}} and {{{y}}} be the two parts of {{{46}}} .
{{{x+y=46}}}
The quotients (if there is no remainder) would be
{{{x/7}}} and {{{y/3}}} such that
{{{x/7+y/3=10}}} ---> {{{3x+7y=210}}}
So, we have the system of linear equations
{{{system(x+y=46,3x+7y=210)}}}-->{{{system(x=46-y,3x+7y=210)}}}-->{{{system(x=46-y,3(46-y)+7y=210)}}}-->{{{system(x=46-y,138-3y+7y=210)}}}-->{{{system(x=46-y,138+4y=210)}}}-->{{{system(x=46-y,4y=210-138)}}}-->{{{system(x=46-y,4y=72)}}}-->{{{system(x=46-y,y=18)}}}-->{{{system(x=46-18,y=18)}}}-->{{{highlight(system(x=28,y=18))}}}
 
NOTE: If the divisions have a quotient and a remainder,
the problem gets complicated.
We would have positive integer quotients {{{a}}} and {{{b}}}
and positive integer remainders {{{R}}} and {{{r}}} .
The two parts of {{{46}}} would be
{{{x=7a+R}}} with {{{1<=R<=6}}}, so {{{R}}}= 1, or 2, or 3, or 4, or 5, or 6, and
{{{y=3b+r}}} with {{{1<=r<=2}}}, so {{{r}}}= 1, or 2.
Then {{{a+b=10}}} and {{{(7a+R)+(3b+r)=46}}}--->{{{7a+3b=46-(R+r)}}} .
We would solve {{{system(a+b=10,7a+3b=46-(R+r))}}} to get
{{{system(a=4-(R+r)/4,b=10-a)}}} .
For {{a}} and {{{b}}} to be integers, with {{{2<=R+r<+6+2=8}}} ,
{{{R+r}}} must be a multiple of {{{4}}}
If {{{R+r=8}}} it must be
{{{system(R=6,r=2,a=4-2=2,b=10-2=8)}}} ---> {{{system(x=7*2+6,y=3*8+2)}}} ---> {{{system(x=20,y=26)}}}
If {{{R+r=4}}} it could be
{{{system(R=2,r=2,a=4-1=3,b=10-3=7)}}} ---> {{{system(x=7*3+2,y=3*7+2)}}} ---> {{{system(x=23,y=23)}}} , or
{{{system(R=3,r=1,a=4-1=3,b=10-3=7)}}} ---> {{{system(x=7*3+3,y=3*7+1)}}} ---> {{{system(x=24,y=22)}}}