Question 121888
Let one of the numbers be n and the other m.


The sum of two numbers is 40, so {{{n+m=40}}}


Let n be the larger number, so twice the larger number is {{{2n}}}, and twice the larger divided by the smaller is {{{2n/m}}}.


The quotient is 3 and the remainder 10 when m is the divisor means {{{3+10/m}}}


Now we have a second equation:


{{{2n/m=3+10/m}}}


Solve the first equation for n by subtracting m from both sides:


{{{n=40-m}}}


Substitute this expression for n into the first equation:


{{{cartoon(2red(n)/m=3+10/m,2*red((40-m))/m=3+10/m)}}}


Multiply both sides by m:


{{{2(40-m)=3m+10}}}


Distribute the 2


{{{80-2m=3m+10}}}


Add -80 and -3m to both sides


{{{-2m-3m=10-80}}}


Collect like terms:


{{{-5m=-70}}}


Divide by -5


{{{m=14}}}


Substitute your value for m into the first equation:


{{{cartoon(n+red(m)=40,n+red(14)=40)}}}
{{{n=26}}}


So the smaller number is 14 and the larger number is 26.